Math  /  Algebra

Questionx25x2x2a2÷x27x+102x25a+2\frac{x^{2}-5 x}{2 x^{2}-a^{2}} \div \frac{x^{2}-7 x+10}{2 x^{2}-5 a+2}

Studdy Solution

STEP 1

Assumptions
1. The given expression is x25x2x2a2÷x27x+102x25a+2\frac{x^{2}-5 x}{2 x^{2}-a^{2}} \div \frac{x^{2}-7 x+10}{2 x^{2}-5 a+2}.
2. We need to simplify the given expression.
3. We assume x0x \neq 0 and a0a \neq 0 to avoid division by zero.

STEP 2

First, rewrite the division as a multiplication by the reciprocal.
x25x2x2a2÷x27x+102x25a+2=x25x2x2a2×2x25a+2x27x+10\frac{x^{2}-5 x}{2 x^{2}-a^{2}} \div \frac{x^{2}-7 x+10}{2 x^{2}-5 a+2} = \frac{x^{2}-5 x}{2 x^{2}-a^{2}} \times \frac{2 x^{2}-5 a+2}{x^{2}-7 x+10}

STEP 3

Factorize the numerators and denominators where possible. Start with the numerator x25xx^{2} - 5x.
x25x=x(x5)x^{2} - 5x = x(x - 5)

STEP 4

Factorize the denominator 2x2a22x^{2} - a^{2} using the difference of squares.
2x2a2=(2xa)(2x+a)2x^{2} - a^{2} = ( \sqrt{2}x - a)( \sqrt{2}x + a)

STEP 5

Factorize the numerator x27x+10x^{2} - 7x + 10.
x27x+10=(x2)(x5)x^{2} - 7x + 10 = (x - 2)(x - 5)

STEP 6

Factorize the denominator 2x25a+22x^{2} - 5a + 2. This expression does not factorize easily, so we leave it as is.
2x25a+22x^{2} - 5a + 2

STEP 7

Substitute the factored forms back into the expression.
x(x5)(2xa)(2x+a)×2x25a+2(x2)(x5)\frac{x(x - 5)}{(\sqrt{2}x - a)(\sqrt{2}x + a)} \times \frac{2x^{2} - 5a + 2}{(x - 2)(x - 5)}

STEP 8

Cancel out the common factors in the numerators and denominators. Here, x5x - 5 is a common factor.
x(2xa)(2x+a)×2x25a+2x2\frac{x}{(\sqrt{2}x - a)(\sqrt{2}x + a)} \times \frac{2x^{2} - 5a + 2}{x - 2}

STEP 9

Combine the remaining fractions.
x(2x25a+2)(2xa)(2x+a)(x2)\frac{x (2x^{2} - 5a + 2)}{(\sqrt{2}x - a)(\sqrt{2}x + a)(x - 2)}
This is the simplified form of the given expression.

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