Simplifying Radical Expressions Involving Fractions worksheets - Free Printable
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Step-by-step solution for: Simplifying Radical Expressions Involving Fractions worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Simplifying Radical Expressions Involving Fractions worksheets
To solve the given problems involving simplifying radical expressions, we will use various techniques such as rationalizing denominators, simplifying fractions with radicals, and combining like terms. Let's go through each problem step by step.
---
Solution:
To simplify this expression, we rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{3} \):
\[
\frac{\sqrt{7}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{7} \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{21}}{3}
\]
Answer:
\[
\boxed{\frac{\sqrt{21}}{3}}
\]
---
Solution:
Similarly, we rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{10} \):
\[
\frac{\sqrt{17}}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}} = \frac{\sqrt{17} \cdot \sqrt{10}}{\sqrt{10} \cdot \sqrt{10}} = \frac{\sqrt{170}}{10}
\]
Answer:
\[
\boxed{\frac{\sqrt{170}}{10}}
\]
---
Solution:
We rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{3} \):
\[
\frac{\sqrt{2}}{5\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{2} \cdot \sqrt{3}}{5\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{6}}{5 \cdot 3} = \frac{\sqrt{6}}{15}
\]
Answer:
\[
\boxed{\frac{\sqrt{6}}{15}}
\]
---
Solution:
We rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{5} \):
\[
\frac{11}{3\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{11 \cdot \sqrt{5}}{3\sqrt{5} \cdot \sqrt{5}} = \frac{11\sqrt{5}}{3 \cdot 5} = \frac{11\sqrt{5}}{15}
\]
Answer:
\[
\boxed{\frac{11\sqrt{5}}{15}}
\]
---
Solution:
First, simplify the denominator \( \sqrt{x^3} \):
\[
\sqrt{x^3} = \sqrt{x^2 \cdot x} = x\sqrt{x}
\]
Now, the expression becomes:
\[
\frac{2\sqrt{3x}}{x\sqrt{x}} = \frac{2\sqrt{3x}}{x\sqrt{x}} = \frac{2\sqrt{3} \cdot \sqrt{x}}{x \cdot \sqrt{x}} = \frac{2\sqrt{3}}{x}
\]
Answer:
\[
\boxed{\frac{2\sqrt{3}}{x}}
\]
---
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{k} \):
\[
\frac{7\sqrt{5}}{\sqrt{k}} \cdot \frac{\sqrt{k}}{\sqrt{k}} = \frac{7\sqrt{5} \cdot \sqrt{k}}{\sqrt{k} \cdot \sqrt{k}} = \frac{7\sqrt{5k}}{k}
\]
Answer:
\[
\boxed{\frac{7\sqrt{5k}}{k}}
\]
---
Solution:
Simplify the numerator and the denominator:
\[
\sqrt{3x^2} = \sqrt{3} \cdot \sqrt{x^2} = \sqrt{3} \cdot x = x\sqrt{3}
\]
\[
\sqrt{5x^2} = \sqrt{5} \cdot \sqrt{x^2} = \sqrt{5} \cdot x = x\sqrt{5}
\]
So the expression becomes:
\[
\frac{x\sqrt{3}}{y^2 \cdot x\sqrt{5}} = \frac{\sqrt{3}}{y^2 \sqrt{5}}
\]
Rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{5} \):
\[
\frac{\sqrt{3}}{y^2 \sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{3} \cdot \sqrt{5}}{y^2 \sqrt{5} \cdot \sqrt{5}} = \frac{\sqrt{15}}{y^2 \cdot 5} = \frac{\sqrt{15}}{5y^2}
\]
Answer:
\[
\boxed{\frac{\sqrt{15}}{5y^2}}
\]
---
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, \( 1 + \sqrt{3} \):
\[
\frac{\sqrt{7}}{1 - \sqrt{3}} \cdot \frac{1 + \sqrt{3}}{1 + \sqrt{3}} = \frac{\sqrt{7}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})}
\]
Simplify the denominator using the difference of squares:
\[
(1 - \sqrt{3})(1 + \sqrt{3}) = 1^2 - (\sqrt{3})^2 = 1 - 3 = -2
\]
So the expression becomes:
\[
\frac{\sqrt{7}(1 + \sqrt{3})}{-2} = \frac{\sqrt{7} + \sqrt{21}}{-2} = -\frac{\sqrt{7} + \sqrt{21}}{2}
\]
Answer:
\[
\boxed{-\frac{\sqrt{7} + \sqrt{21}}{2}}
\]
---
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, \( \sqrt{a} + \sqrt{b} \):
\[
\frac{\sqrt{a}}{\sqrt{a} - \sqrt{b}} \cdot \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}} = \frac{\sqrt{a}(\sqrt{a} + \sqrt{b})}{(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b})}
\]
Simplify the denominator using the difference of squares:
\[
(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b
\]
So the expression becomes:
\[
\frac{\sqrt{a}(\sqrt{a} + \sqrt{b})}{a - b} = \frac{a + \sqrt{ab}}{a - b}
\]
Answer:
\[
\boxed{\frac{a + \sqrt{ab}}{a - b}}
\]
---
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, \( 2 - \sqrt{5} \):
\[
\frac{1 - \sqrt{3}}{2 + \sqrt{5}} \cdot \frac{2 - \sqrt{5}}{2 - \sqrt{5}} = \frac{(1 - \sqrt{3})(2 - \sqrt{5})}{(2 + \sqrt{5})(2 - \sqrt{5})}
\]
Simplify the denominator using the difference of squares:
\[
(2 + \sqrt{5})(2 - \sqrt{5}) = 2^2 - (\sqrt{5})^2 = 4 - 5 = -1
\]
Expand the numerator:
\[
(1 - \sqrt{3})(2 - \sqrt{5}) = 1 \cdot 2 - 1 \cdot \sqrt{5} - \sqrt{3} \cdot 2 + \sqrt{3} \cdot \sqrt{5} = 2 - \sqrt{5} - 2\sqrt{3} + \sqrt{15}
\]
So the expression becomes:
\[
\frac{2 - \sqrt{5} - 2\sqrt{3} + \sqrt{15}}{-1} = -2 + \sqrt{5} + 2\sqrt{3} - \sqrt{15}
\]
Answer:
\[
\boxed{-2 + \sqrt{5} + 2\sqrt{3} - \sqrt{15}}
\]
---
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, \( 1 + \sqrt{2} \):
\[
\frac{1}{1 - \sqrt{2}} \cdot \frac{1 + \sqrt{2}}{1 + \sqrt{2}} = \frac{1(1 + \sqrt{2})}{(1 - \sqrt{2})(1 + \sqrt{2})}
\]
Simplify the denominator using the difference of squares:
\[
(1 - \sqrt{2})(1 + \sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1 - 2 = -1
\]
So the expression becomes:
\[
\frac{1 + \sqrt{2}}{-1} = -1 - \sqrt{2}
\]
Answer:
\[
\boxed{-1 - \sqrt{2}}
\]
---
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, \( 5 - \sqrt{3} \):
\[
\frac{7}{5 + \sqrt{3}} \cdot \frac{5 - \sqrt{3}}{5 - \sqrt{3}} = \frac{7(5 - \sqrt{3})}{(5 + \sqrt{3})(5 - \sqrt{3})}
\]
Simplify the denominator using the difference of squares:
\[
(5 + \sqrt{3})(5 - \sqrt{3}) = 5^2 - (\sqrt{3})^2 = 25 - 3 = 22
\]
Expand the numerator:
\[
7(5 - \sqrt{3}) = 7 \cdot 5 - 7 \cdot \sqrt{3} = 35 - 7\sqrt{3}
\]
So the expression becomes:
\[
\frac{35 - 7\sqrt{3}}{22}
\]
Answer:
\[
\boxed{\frac{35 - 7\sqrt{3}}{22}}
\]
---
Solution:
Notice that the numerator and the denominator are very similar but have opposite signs. We can factor out a negative sign from the denominator:
\[
\frac{\sqrt{5} - \sqrt{3}}{\sqrt{3} - \sqrt{5}} = \frac{\sqrt{5} - \sqrt{3}}{-(\sqrt{5} - \sqrt{3})} = -1
\]
Answer:
\[
\boxed{-1}
\]
---
Solution:
This expression is already in its simplest form because the numerator and the denominator do not share any common factors that can be simplified further. The terms in the numerator and the denominator are not conjugates, and no further rationalization or simplification is possible.
Answer:
\[
\boxed{\frac{2\sqrt{2} - 7\sqrt{5}}{\sqrt{5} - 3\sqrt{2}}}
\]
---
Solution:
This expression is already in its simplest form because the terms in the numerator and the denominator do not share any common factors that can be simplified further. No further rationalization or simplification is possible.
Answer:
\[
\boxed{\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{5}}}
\]
---
Solution:
Simplify the numerator and the denominator separately:
\[
\sqrt{y^3x^7} = \sqrt{y^2 \cdot y \cdot x^6 \cdot x} = y \cdot x^3 \cdot \sqrt{yx}
\]
\[
\sqrt{5y^5x^3} = \sqrt{5 \cdot y^4 \cdot y \cdot x^2 \cdot x} = y^2 \cdot x \cdot \sqrt{5yx}
\]
So the expression becomes:
\[
\frac{y \cdot x^3 \cdot \sqrt{yx}}{y^2 \cdot x \cdot \sqrt{5yx}} = \frac{y \cdot x^3}{y^2 \cdot x} \cdot \frac{\sqrt{yx}}{\sqrt{5yx}} = \frac{x^2}{y} \cdot \frac{1}{\sqrt{5}} = \frac{x^2}{y\sqrt{5}}
\]
Rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{5} \):
\[
\frac{x^2}{y\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{x^2 \sqrt{5}}{y \cdot 5} = \frac{x^2 \sqrt{5}}{5y}
\]
Answer:
\[
\boxed{\frac{x^2 \sqrt{5}}{5y}}
\]
---
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{6} \):
\[
\frac{\sqrt{11}}{3\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{11} \cdot \sqrt{6}}{3\sqrt{6} \cdot \sqrt{6}} = \frac{\sqrt{66}}{3 \cdot 6} = \frac{\sqrt{66}}{18}
\]
Answer:
\[
\boxed{\frac{\sqrt{66}}{18}}
\]
---
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{y} \):
\[
\frac{m\sqrt{n}}{\sqrt{y}} \cdot \frac{\sqrt{y}}{\sqrt{y}} = \frac{m\sqrt{n} \cdot \sqrt{y}}{\sqrt{y} \cdot \sqrt{y}} = \frac{m\sqrt{ny}}{y}
\]
Answer:
\[
\boxed{\frac{m\sqrt{ny}}{y}}
\]
---
\[
\boxed{
\begin{aligned}
1. & \ \frac{\sqrt{21}}{3} \\
2. & \ \frac{\sqrt{170}}{10} \\
3. & \ \frac{\sqrt{6}}{15} \\
4. & \ \frac{11\sqrt{5}}{15} \\
5. & \ \frac{2\sqrt{3}}{x} \\
6. & \ \frac{7\sqrt{5k}}{k} \\
7. & \ \frac{\sqrt{15}}{5y^2} \\
8. & \ -\frac{\sqrt{7} + \sqrt{21}}{2} \\
9. & \ \frac{a + \sqrt{ab}}{a - b} \\
10. & \ -2 + \sqrt{5} + 2\sqrt{3} - \sqrt{15} \\
11. & \ -1 - \sqrt{2} \\
12. & \ \frac{35 - 7\sqrt{3}}{22} \\
13. & \ -1 \\
14. & \ \frac{2\sqrt{2} - 7\sqrt{5}}{\sqrt{5} - 3\sqrt{2}} \\
15. & \ \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{5}} \\
16. & \ \frac{x^2 \sqrt{5}}{5y} \\
17. & \ \frac{\sqrt{66}}{18} \\
18. & \ \frac{m\sqrt{ny}}{y}
\end{aligned}
}
\]
---
Problem 1: Simplify \( \frac{\sqrt{7}}{\sqrt{3}} \)
Solution:
To simplify this expression, we rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{3} \):
\[
\frac{\sqrt{7}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{7} \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{21}}{3}
\]
Answer:
\[
\boxed{\frac{\sqrt{21}}{3}}
\]
---
Problem 2: Simplify \( \frac{\sqrt{17}}{\sqrt{10}} \)
Solution:
Similarly, we rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{10} \):
\[
\frac{\sqrt{17}}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}} = \frac{\sqrt{17} \cdot \sqrt{10}}{\sqrt{10} \cdot \sqrt{10}} = \frac{\sqrt{170}}{10}
\]
Answer:
\[
\boxed{\frac{\sqrt{170}}{10}}
\]
---
Problem 3: Simplify \( \frac{\sqrt{2}}{5\sqrt{3}} \)
Solution:
We rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{3} \):
\[
\frac{\sqrt{2}}{5\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{2} \cdot \sqrt{3}}{5\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{6}}{5 \cdot 3} = \frac{\sqrt{6}}{15}
\]
Answer:
\[
\boxed{\frac{\sqrt{6}}{15}}
\]
---
Problem 4: Simplify \( \frac{11}{3\sqrt{5}} \)
Solution:
We rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{5} \):
\[
\frac{11}{3\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{11 \cdot \sqrt{5}}{3\sqrt{5} \cdot \sqrt{5}} = \frac{11\sqrt{5}}{3 \cdot 5} = \frac{11\sqrt{5}}{15}
\]
Answer:
\[
\boxed{\frac{11\sqrt{5}}{15}}
\]
---
Problem 5: Simplify \( \frac{2\sqrt{3x}}{\sqrt{x^3}} \)
Solution:
First, simplify the denominator \( \sqrt{x^3} \):
\[
\sqrt{x^3} = \sqrt{x^2 \cdot x} = x\sqrt{x}
\]
Now, the expression becomes:
\[
\frac{2\sqrt{3x}}{x\sqrt{x}} = \frac{2\sqrt{3x}}{x\sqrt{x}} = \frac{2\sqrt{3} \cdot \sqrt{x}}{x \cdot \sqrt{x}} = \frac{2\sqrt{3}}{x}
\]
Answer:
\[
\boxed{\frac{2\sqrt{3}}{x}}
\]
---
Problem 6: Simplify \( \frac{7\sqrt{5}}{\sqrt{k}} \)
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{k} \):
\[
\frac{7\sqrt{5}}{\sqrt{k}} \cdot \frac{\sqrt{k}}{\sqrt{k}} = \frac{7\sqrt{5} \cdot \sqrt{k}}{\sqrt{k} \cdot \sqrt{k}} = \frac{7\sqrt{5k}}{k}
\]
Answer:
\[
\boxed{\frac{7\sqrt{5k}}{k}}
\]
---
Problem 7: Simplify \( \frac{\sqrt{3x^2}}{y^2\sqrt{5x^2}} \)
Solution:
Simplify the numerator and the denominator:
\[
\sqrt{3x^2} = \sqrt{3} \cdot \sqrt{x^2} = \sqrt{3} \cdot x = x\sqrt{3}
\]
\[
\sqrt{5x^2} = \sqrt{5} \cdot \sqrt{x^2} = \sqrt{5} \cdot x = x\sqrt{5}
\]
So the expression becomes:
\[
\frac{x\sqrt{3}}{y^2 \cdot x\sqrt{5}} = \frac{\sqrt{3}}{y^2 \sqrt{5}}
\]
Rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{5} \):
\[
\frac{\sqrt{3}}{y^2 \sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{3} \cdot \sqrt{5}}{y^2 \sqrt{5} \cdot \sqrt{5}} = \frac{\sqrt{15}}{y^2 \cdot 5} = \frac{\sqrt{15}}{5y^2}
\]
Answer:
\[
\boxed{\frac{\sqrt{15}}{5y^2}}
\]
---
Problem 8: Simplify \( \frac{\sqrt{7}}{1 - \sqrt{3}} \)
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, \( 1 + \sqrt{3} \):
\[
\frac{\sqrt{7}}{1 - \sqrt{3}} \cdot \frac{1 + \sqrt{3}}{1 + \sqrt{3}} = \frac{\sqrt{7}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})}
\]
Simplify the denominator using the difference of squares:
\[
(1 - \sqrt{3})(1 + \sqrt{3}) = 1^2 - (\sqrt{3})^2 = 1 - 3 = -2
\]
So the expression becomes:
\[
\frac{\sqrt{7}(1 + \sqrt{3})}{-2} = \frac{\sqrt{7} + \sqrt{21}}{-2} = -\frac{\sqrt{7} + \sqrt{21}}{2}
\]
Answer:
\[
\boxed{-\frac{\sqrt{7} + \sqrt{21}}{2}}
\]
---
Problem 9: Simplify \( \frac{\sqrt{a}}{\sqrt{a} - \sqrt{b}} \)
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, \( \sqrt{a} + \sqrt{b} \):
\[
\frac{\sqrt{a}}{\sqrt{a} - \sqrt{b}} \cdot \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}} = \frac{\sqrt{a}(\sqrt{a} + \sqrt{b})}{(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b})}
\]
Simplify the denominator using the difference of squares:
\[
(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b
\]
So the expression becomes:
\[
\frac{\sqrt{a}(\sqrt{a} + \sqrt{b})}{a - b} = \frac{a + \sqrt{ab}}{a - b}
\]
Answer:
\[
\boxed{\frac{a + \sqrt{ab}}{a - b}}
\]
---
Problem 10: Simplify \( \frac{1 - \sqrt{3}}{2 + \sqrt{5}} \)
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, \( 2 - \sqrt{5} \):
\[
\frac{1 - \sqrt{3}}{2 + \sqrt{5}} \cdot \frac{2 - \sqrt{5}}{2 - \sqrt{5}} = \frac{(1 - \sqrt{3})(2 - \sqrt{5})}{(2 + \sqrt{5})(2 - \sqrt{5})}
\]
Simplify the denominator using the difference of squares:
\[
(2 + \sqrt{5})(2 - \sqrt{5}) = 2^2 - (\sqrt{5})^2 = 4 - 5 = -1
\]
Expand the numerator:
\[
(1 - \sqrt{3})(2 - \sqrt{5}) = 1 \cdot 2 - 1 \cdot \sqrt{5} - \sqrt{3} \cdot 2 + \sqrt{3} \cdot \sqrt{5} = 2 - \sqrt{5} - 2\sqrt{3} + \sqrt{15}
\]
So the expression becomes:
\[
\frac{2 - \sqrt{5} - 2\sqrt{3} + \sqrt{15}}{-1} = -2 + \sqrt{5} + 2\sqrt{3} - \sqrt{15}
\]
Answer:
\[
\boxed{-2 + \sqrt{5} + 2\sqrt{3} - \sqrt{15}}
\]
---
Problem 11: Simplify \( \frac{1}{1 - \sqrt{2}} \)
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, \( 1 + \sqrt{2} \):
\[
\frac{1}{1 - \sqrt{2}} \cdot \frac{1 + \sqrt{2}}{1 + \sqrt{2}} = \frac{1(1 + \sqrt{2})}{(1 - \sqrt{2})(1 + \sqrt{2})}
\]
Simplify the denominator using the difference of squares:
\[
(1 - \sqrt{2})(1 + \sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1 - 2 = -1
\]
So the expression becomes:
\[
\frac{1 + \sqrt{2}}{-1} = -1 - \sqrt{2}
\]
Answer:
\[
\boxed{-1 - \sqrt{2}}
\]
---
Problem 12: Simplify \( \frac{7}{5 + \sqrt{3}} \)
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, \( 5 - \sqrt{3} \):
\[
\frac{7}{5 + \sqrt{3}} \cdot \frac{5 - \sqrt{3}}{5 - \sqrt{3}} = \frac{7(5 - \sqrt{3})}{(5 + \sqrt{3})(5 - \sqrt{3})}
\]
Simplify the denominator using the difference of squares:
\[
(5 + \sqrt{3})(5 - \sqrt{3}) = 5^2 - (\sqrt{3})^2 = 25 - 3 = 22
\]
Expand the numerator:
\[
7(5 - \sqrt{3}) = 7 \cdot 5 - 7 \cdot \sqrt{3} = 35 - 7\sqrt{3}
\]
So the expression becomes:
\[
\frac{35 - 7\sqrt{3}}{22}
\]
Answer:
\[
\boxed{\frac{35 - 7\sqrt{3}}{22}}
\]
---
Problem 13: Simplify \( \frac{\sqrt{5} - \sqrt{3}}{\sqrt{3} - \sqrt{5}} \)
Solution:
Notice that the numerator and the denominator are very similar but have opposite signs. We can factor out a negative sign from the denominator:
\[
\frac{\sqrt{5} - \sqrt{3}}{\sqrt{3} - \sqrt{5}} = \frac{\sqrt{5} - \sqrt{3}}{-(\sqrt{5} - \sqrt{3})} = -1
\]
Answer:
\[
\boxed{-1}
\]
---
Problem 14: Simplify \( \frac{2\sqrt{2} - 7\sqrt{5}}{\sqrt{5} - 3\sqrt{2}} \)
Solution:
This expression is already in its simplest form because the numerator and the denominator do not share any common factors that can be simplified further. The terms in the numerator and the denominator are not conjugates, and no further rationalization or simplification is possible.
Answer:
\[
\boxed{\frac{2\sqrt{2} - 7\sqrt{5}}{\sqrt{5} - 3\sqrt{2}}}
\]
---
Problem 15: Simplify \( \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{5}} \)
Solution:
This expression is already in its simplest form because the terms in the numerator and the denominator do not share any common factors that can be simplified further. No further rationalization or simplification is possible.
Answer:
\[
\boxed{\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{5}}}
\]
---
Problem 16: Simplify \( \frac{\sqrt{y^3x^7}}{\sqrt{5y^5x^3}} \)
Solution:
Simplify the numerator and the denominator separately:
\[
\sqrt{y^3x^7} = \sqrt{y^2 \cdot y \cdot x^6 \cdot x} = y \cdot x^3 \cdot \sqrt{yx}
\]
\[
\sqrt{5y^5x^3} = \sqrt{5 \cdot y^4 \cdot y \cdot x^2 \cdot x} = y^2 \cdot x \cdot \sqrt{5yx}
\]
So the expression becomes:
\[
\frac{y \cdot x^3 \cdot \sqrt{yx}}{y^2 \cdot x \cdot \sqrt{5yx}} = \frac{y \cdot x^3}{y^2 \cdot x} \cdot \frac{\sqrt{yx}}{\sqrt{5yx}} = \frac{x^2}{y} \cdot \frac{1}{\sqrt{5}} = \frac{x^2}{y\sqrt{5}}
\]
Rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{5} \):
\[
\frac{x^2}{y\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{x^2 \sqrt{5}}{y \cdot 5} = \frac{x^2 \sqrt{5}}{5y}
\]
Answer:
\[
\boxed{\frac{x^2 \sqrt{5}}{5y}}
\]
---
Problem 17: Simplify \( \frac{\sqrt{11}}{3\sqrt{6}} \)
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{6} \):
\[
\frac{\sqrt{11}}{3\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{11} \cdot \sqrt{6}}{3\sqrt{6} \cdot \sqrt{6}} = \frac{\sqrt{66}}{3 \cdot 6} = \frac{\sqrt{66}}{18}
\]
Answer:
\[
\boxed{\frac{\sqrt{66}}{18}}
\]
---
Problem 18: Simplify \( \frac{m\sqrt{n}}{\sqrt{y}} \)
Solution:
Rationalize the denominator by multiplying both the numerator and the denominator by \( \sqrt{y} \):
\[
\frac{m\sqrt{n}}{\sqrt{y}} \cdot \frac{\sqrt{y}}{\sqrt{y}} = \frac{m\sqrt{n} \cdot \sqrt{y}}{\sqrt{y} \cdot \sqrt{y}} = \frac{m\sqrt{ny}}{y}
\]
Answer:
\[
\boxed{\frac{m\sqrt{ny}}{y}}
\]
---
Final Answer:
\[
\boxed{
\begin{aligned}
1. & \ \frac{\sqrt{21}}{3} \\
2. & \ \frac{\sqrt{170}}{10} \\
3. & \ \frac{\sqrt{6}}{15} \\
4. & \ \frac{11\sqrt{5}}{15} \\
5. & \ \frac{2\sqrt{3}}{x} \\
6. & \ \frac{7\sqrt{5k}}{k} \\
7. & \ \frac{\sqrt{15}}{5y^2} \\
8. & \ -\frac{\sqrt{7} + \sqrt{21}}{2} \\
9. & \ \frac{a + \sqrt{ab}}{a - b} \\
10. & \ -2 + \sqrt{5} + 2\sqrt{3} - \sqrt{15} \\
11. & \ -1 - \sqrt{2} \\
12. & \ \frac{35 - 7\sqrt{3}}{22} \\
13. & \ -1 \\
14. & \ \frac{2\sqrt{2} - 7\sqrt{5}}{\sqrt{5} - 3\sqrt{2}} \\
15. & \ \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{5}} \\
16. & \ \frac{x^2 \sqrt{5}}{5y} \\
17. & \ \frac{\sqrt{66}}{18} \\
18. & \ \frac{m\sqrt{ny}}{y}
\end{aligned}
}
\]
Parent Tip: Review the logic above to help your child master the concept of simplifying radicals worksheet algebra 1.
