Math Problem Statement

Simplify the expression \( \frac{5\sqrt{2}}{\sqrt{10}-\sqrt{5}} - \frac{3\sqrt{5}}{\sqrt{5}-\sqrt{2}} - \frac{\sqrt{10}}{\sqrt{2}-1} \).

Solution

The given expression is:

5210535521021\frac{5 \sqrt{2}}{\sqrt{10} - \sqrt{5}} - \frac{3 \sqrt{5}}{\sqrt{5} - \sqrt{2}} - \frac{\sqrt{10}}{\sqrt{2} - 1}

To simplify this, we need to rationalize the denominators of each fraction. Let's handle each term individually.

Step 1: Rationalize 52105\frac{5 \sqrt{2}}{\sqrt{10} - \sqrt{5}}

We multiply both the numerator and denominator by the conjugate of the denominator, 10+5\sqrt{10} + \sqrt{5}.

5210510+510+5=52(10+5)(10)2(5)2\frac{5 \sqrt{2}}{\sqrt{10} - \sqrt{5}} \cdot \frac{\sqrt{10} + \sqrt{5}}{\sqrt{10} + \sqrt{5}} = \frac{5 \sqrt{2}(\sqrt{10} + \sqrt{5})}{(\sqrt{10})^2 - (\sqrt{5})^2}

The denominator simplifies to:

(10)2(5)2=105=5(\sqrt{10})^2 - (\sqrt{5})^2 = 10 - 5 = 5

Now the expression becomes:

52(10+5)5=2(10+5)\frac{5 \sqrt{2}(\sqrt{10} + \sqrt{5})}{5} = \sqrt{2}(\sqrt{10} + \sqrt{5})

Distribute the 2\sqrt{2}:

210=20=25,25=10\sqrt{2} \cdot \sqrt{10} = \sqrt{20} = 2 \sqrt{5}, \quad \sqrt{2} \cdot \sqrt{5} = \sqrt{10}

So, the first term becomes:

25+102 \sqrt{5} + \sqrt{10}

Step 2: Rationalize 3552\frac{3 \sqrt{5}}{\sqrt{5} - \sqrt{2}}

We multiply both the numerator and denominator by the conjugate of the denominator, 5+2\sqrt{5} + \sqrt{2}:

35525+25+2=35(5+2)(5)2(2)2\frac{3 \sqrt{5}}{\sqrt{5} - \sqrt{2}} \cdot \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}} = \frac{3 \sqrt{5}(\sqrt{5} + \sqrt{2})}{(\sqrt{5})^2 - (\sqrt{2})^2}

The denominator simplifies to:

(5)2(2)2=52=3(\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3

Now the expression becomes:

35(5+2)3=5(5+2)\frac{3 \sqrt{5}(\sqrt{5} + \sqrt{2})}{3} = \sqrt{5}(\sqrt{5} + \sqrt{2})

Distribute the 5\sqrt{5}:

55=5,52=10\sqrt{5} \cdot \sqrt{5} = 5, \quad \sqrt{5} \cdot \sqrt{2} = \sqrt{10}

So, the second term becomes:

5+105 + \sqrt{10}

Step 3: Rationalize 1021\frac{\sqrt{10}}{\sqrt{2} - 1}

We multiply both the numerator and denominator by the conjugate of the denominator, 2+1\sqrt{2} + 1:

10212+12+1=10(2+1)(2)212\frac{\sqrt{10}}{\sqrt{2} - 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{\sqrt{10}(\sqrt{2} + 1)}{(\sqrt{2})^2 - 1^2}

The denominator simplifies to:

(2)212=21=1(\sqrt{2})^2 - 1^2 = 2 - 1 = 1

So, the third term becomes:

10(2+1)=102+10=20+10=25+10\sqrt{10}(\sqrt{2} + 1) = \sqrt{10} \cdot \sqrt{2} + \sqrt{10} = \sqrt{20} + \sqrt{10} = 2 \sqrt{5} + \sqrt{10}

Step 4: Combine the terms

We now have:

(25+10)(5+10)(25+10)(2 \sqrt{5} + \sqrt{10}) - (5 + \sqrt{10}) - (2 \sqrt{5} + \sqrt{10})

Simplify by combining like terms:

  1. The 10\sqrt{10} terms cancel out since +101010=0+\sqrt{10} - \sqrt{10} - \sqrt{10} = 0.
  2. The 252 \sqrt{5} terms cancel out as well: 2525=02 \sqrt{5} - 2 \sqrt{5} = 0.

This leaves:

05=50 - 5 = -5

Thus, the simplified expression is:

5\boxed{-5}

Would you like more details on any step? Here are some related questions to expand upon:

  1. What are conjugates, and how are they useful in rationalizing denominators?
  2. Why do we rationalize the denominator, and how does it affect expressions?
  3. Can you show examples of simplifying expressions with other radicals?
  4. How does one simplify the product of two square roots?
  5. How do you approach more complex radical expressions?

Tip: Always look for common terms that can cancel out when simplifying radical expressions to make calculations easier.

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Math Problem Analysis

Mathematical Concepts

Radical Expressions
Rationalization
Simplification of Algebraic Expressions

Formulas

Conjugate multiplication: \( (a - b)(a + b) = a^2 - b^2 \)
Square root simplification: \( \sqrt{a} \times \sqrt{b} = \sqrt{ab} \)

Theorems

Simplification of Radicals
Rationalizing the Denominator

Suitable Grade Level

Grades 10-12