Math worksheet for practicing rationalizing denominators with various algebraic expressions.

Let’s solve each problem one by one. We’ll rationalize the denominator — that means we get rid of any square roots in the bottom part (denominator) of the fraction.

---

Problem 1:
We have:
\[
-\frac{\sqrt{5} - 2}{\sqrt{5} - 12}
\]

To rationalize, multiply numerator and denominator by the conjugate of the denominator. The conjugate of \(\sqrt{5} - 12\) is \(\sqrt{5} + 12\).

So:

Numerator:
\[
-(\sqrt{5} - 2)(\sqrt{5} + 12) = -[\sqrt{5}\cdot\sqrt{5} + 12\sqrt{5} - 2\sqrt{5} - 24] = -[5 + 10\sqrt{5} - 24] = -[-19 + 10\sqrt{5}] = 19 - 10\sqrt{5}
\]

Denominator:
\[
(\sqrt{5} - 12)(\sqrt{5} + 12) = (\sqrt{5})^2 - (12)^2 = 5 - 144 = -139
\]

So the whole thing becomes:
\[
\frac{19 - 10\sqrt{5}}{-139} = -\frac{19 - 10\sqrt{5}}{139} = \frac{-19 + 10\sqrt{5}}{139}
\]

✔ Final Answer for #1: \(\boxed{\frac{-19 + 10\sqrt{5}}{139}}\)

---

Problem 2:
\[
\frac{4\sqrt{6}}{\sqrt{32}}
\]

First, simplify \(\sqrt{32}\):
\(\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}\)

So now:
\[
\frac{4\sqrt{6}}{4\sqrt{2}} = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{\frac{6}{2}} = \sqrt{3}
\]

✔ Final Answer for #2: \(\boxed{\sqrt{3}}\)

---

Problem 3:
\[
\frac{8\sqrt{2}}{\sqrt{16}}
\]

\(\sqrt{16} = 4\), so:
\[
\frac{8\sqrt{2}}{4} = 2\sqrt{2}
\]

✔ Final Answer for #3: \(\boxed{2\sqrt{2}}\)

---

Problem 4:
\[
\frac{\sqrt{2} + 10}{\sqrt{2} - 10}
\]

Multiply numerator and denominator by conjugate of denominator: \(\sqrt{2} + 10\)

Numerator:
\[
(\sqrt{2} + 10)(\sqrt{2} + 10) = (\sqrt{2})^2 + 2\cdot\sqrt{2}\cdot10 + 10^2 = 2 + 20\sqrt{2} + 100 = 102 + 20\sqrt{2}
\]

Denominator:
\[
(\sqrt{2} - 10)(\sqrt{2} + 10) = (\sqrt{2})^2 - (10)^2 = 2 - 100 = -98
\]

So:
\[
\frac{102 + 20\sqrt{2}}{-98} = -\frac{102 + 20\sqrt{2}}{98}
\]

Simplify by dividing numerator and denominator by 2:
\[
-\frac{51 + 10\sqrt{2}}{49}
\]

✔ Final Answer for #4: \(\boxed{-\frac{51 + 10\sqrt{2}}{49}}\)

---

Problem 5:
\[
\frac{14 + \sqrt{7}}{4\sqrt{7} + 11}
\]

Conjugate of denominator \(4\sqrt{7} + 11\) is \(4\sqrt{7} - 11\)

Multiply numerator and denominator by that.

Numerator:
\[
(14 + \sqrt{7})(4\sqrt{7} - 11) = 14\cdot4\sqrt{7} - 14\cdot11 + \sqrt{7}\cdot4\sqrt{7} - \sqrt{7}\cdot11
= 56\sqrt{7} - 154 + 4\cdot7 - 11\sqrt{7}
= 56\sqrt{7} - 154 + 28 - 11\sqrt{7}
= (56\sqrt{7} - 11\sqrt{7}) + (-154 + 28)
= 45\sqrt{7} - 126
\]

Denominator:
\[
(4\sqrt{7} + 11)(4\sqrt{7} - 11) = (4\sqrt{7})^2 - (11)^2 = 16\cdot7 - 121 = 112 - 121 = -9
\]

So:
\[
\frac{45\sqrt{7} - 126}{-9} = -\frac{45\sqrt{7} - 126}{9} = \frac{-45\sqrt{7} + 126}{9} = \frac{126 - 45\sqrt{7}}{9}
\]

Divide numerator and denominator by 3:
\[
\frac{42 - 15\sqrt{7}}{3}
\]

Wait — actually, let’s divide each term by 9 directly:

\[
\frac{45\sqrt{7}}{-9} = -5\sqrt{7}, \quad \frac{-126}{-9} = 14
\Rightarrow 14 - 5\sqrt{7}
\]

Oh! I made a sign error earlier. Let me fix:

Original after multiplying:
Numerator: \(45\sqrt{7} - 126\)
Denominator: \(-9\)

So:
\[
\frac{45\sqrt{7} - 126}{-9} = \frac{45\sqrt{7}}{-9} + \frac{-126}{-9} = -5\sqrt{7} + 14 = 14 - 5\sqrt{7}
\]

✔ Final Answer for #5: \(\boxed{14 - 5\sqrt{7}}\)

---

Problem 6:
\[
-\frac{15}{1 - \sqrt{19}}
\]

Multiply numerator and denominator by conjugate: \(1 + \sqrt{19}\)

Numerator:
\[
-15(1 + \sqrt{19}) = -15 - 15\sqrt{19}
\]

Denominator:
\[
(1 - \sqrt{19})(1 + \sqrt{19}) = 1^2 - (\sqrt{19})^2 = 1 - 19 = -18
\]

So:
\[
\frac{-15 - 15\sqrt{19}}{-18} = \frac{15 + 15\sqrt{19}}{18}
\]

Factor out 15:
\[
\frac{15(1 + \sqrt{19})}{18} = \frac{5(1 + \sqrt{19})}{6}
\]

✔ Final Answer for #6: \(\boxed{\frac{5(1 + \sqrt{19})}{6}}\)

---

Problem 7:
\[
\frac{9\sqrt{2} - 8}{4 - 3\sqrt{3}}
\]

Conjugate of denominator \(4 - 3\sqrt{3}\) is \(4 + 3\sqrt{3}\)

Multiply numerator and denominator by that.

Numerator:
\[
(9\sqrt{2} - 8)(4 + 3\sqrt{3}) = 9\sqrt{2}\cdot4 + 9\sqrt{2}\cdot3\sqrt{3} -8\cdot4 -8\cdot3\sqrt{3}
= 36\sqrt{2} + 27\sqrt{6} - 32 - 24\sqrt{3}
\]

Denominator:
\[
(4 - 3\sqrt{3})(4 + 3\sqrt{3}) = 4^2 - (3\sqrt{3})^2 = 16 - 9\cdot3 = 16 - 27 = -11
\]

So:
\[
\frac{36\sqrt{2} + 27\sqrt{6} - 32 - 24\sqrt{3}}{-11} = -\frac{36\sqrt{2} + 27\sqrt{6} - 32 - 24\sqrt{3}}{11}
\]

Or write as:
\[
\frac{-36\sqrt{2} - 27\sqrt{6} + 32 + 24\sqrt{3}}{11}
\]

✔ Final Answer for #7: \(\boxed{\frac{32 + 24\sqrt{3} - 36\sqrt{2} - 27\sqrt{6}}{11}}\)

---

Problem 8:
\[
\frac{\sqrt{2} - \sqrt{10}}{\sqrt{15} + \sqrt{12}}
\]

First, simplify \(\sqrt{12} = 2\sqrt{3}\), so denominator is \(\sqrt{15} + 2\sqrt{3}\)

But it’s easier to rationalize with conjugate: \(\sqrt{15} - \sqrt{12}\)

Multiply numerator and denominator by \(\sqrt{15} - \sqrt{12}\)

Numerator:
\[
(\sqrt{2} - \sqrt{10})(\sqrt{15} - \sqrt{12}) = \sqrt{2}\cdot\sqrt{15} - \sqrt{2}\cdot\sqrt{12} - \sqrt{10}\cdot\sqrt{15} + \sqrt{10}\cdot\sqrt{12}
= \sqrt{30} - \sqrt{24} - \sqrt{150} + \sqrt{120}
\]

Simplify radicals:

- \(\sqrt{24} = 2\sqrt{6}\)
- \(\sqrt{150} = \sqrt{25\cdot6} = 5\sqrt{6}\)
- \(\sqrt{120} = \sqrt{4\cdot30} = 2\sqrt{30}\)

So numerator becomes:
\[
\sqrt{30} - 2\sqrt{6} - 5\sqrt{6} + 2\sqrt{30} = (1+2)\sqrt{30} + (-2-5)\sqrt{6} = 3\sqrt{30} - 7\sqrt{6}
\]

Denominator:
\[
(\sqrt{15} + \sqrt{12})(\sqrt{15} - \sqrt{12}) = (\sqrt{15})^2 - (\sqrt{12})^2 = 15 - 12 = 3
\]

So:
\[
\frac{3\sqrt{30} - 7\sqrt{6}}{3} = \sqrt{30} - \frac{7\sqrt{6}}{3}
\]

Wait — better to leave as single fraction:

\[
\frac{3\sqrt{30} - 7\sqrt{6}}{3}
\]

✔ Final Answer for #8: \(\boxed{\frac{3\sqrt{30} - 7\sqrt{6}}{3}}\)

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Final Answers:

1) \(\boxed{\frac{-19 + 10\sqrt{5}}{139}}\)
2) \(\boxed{\sqrt{3}}\)
3) \(\boxed{2\sqrt{2}}\)
4) \(\boxed{-\frac{51 + 10\sqrt{2}}{49}}\)
5) \(\boxed{14 - 5\sqrt{7}}\)
6) \(\boxed{\frac{5(1 + \sqrt{19})}{6}}\)
7) \(\boxed{\frac{32 + 24\sqrt{3} - 36\sqrt{2} - 27\sqrt{6}}{11}}\)
8) \(\boxed{\frac{3\sqrt{30} - 7\sqrt{6}}{3}}\)