Question

RÚT GỌN

a) A= (3x - x2/ 9 - x2 - 1 ) : ( 9-x2 / x2 + x + 6 - x-3/ 2-x - x+2/ x+3

b) B= (x+2/ x-1 - x+1/ x+2 + x+5/x2 + 2 - 2) . ( x3-1)

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Answer 1

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Answer 2

a: \(\left(\dfrac{3x-x^2}{9-x^2}-1\right):\left(\dfrac{9-x^2}{x^2+x-6}-\dfrac{x-3}{2-x}-\dfrac{x+2}{x+3}\right)\)

\(=\left(\dfrac{x}{x+3}-1\right):\dfrac{9-x^2+x^2-9-x^2+4}{\left(x-2\right)\left(x+3\right)}\)

\(=\dfrac{x-x-3}{x+3}\cdot\dfrac{\left(x+3\right)\left(x-2\right)}{-x^2+4}\)

\(=\dfrac{-3}{1}\cdot\dfrac{\left(x-2\right)}{-\left(x-2\right)\left(x+2\right)}=\dfrac{3}{x+2}\)

b: \(\left(\dfrac{x+2}{x-1}-\dfrac{x-1}{x+2}+\dfrac{x+5}{x^2+x-2}\right)\cdot\left(x^3-1\right)\)

\(=\dfrac{x^2+4x+4-x^2+2x-1+x+5}{\left(x+2\right)\left(x-1\right)}\cdot\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{1}\)

\(=\dfrac{7x+8}{\left(x+2\right)}\cdot\dfrac{x^2+x+1}{1}\)