Question

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Answer 1

a)ĐKXĐ:\(x\ne0,2\)

\(A=\left(\dfrac{-x}{2-x}+\dfrac{3x-2}{x\left(2-x\right)}\right):\left(\dfrac{\left(x+2\right)\left(x-1\right)+\left(4-x\right)x}{x\left(x-2\right)}\right)=\left(\dfrac{-x^2}{x\left(2-x\right)}+\dfrac{3x-2}{x\left(2-x\right)}\right):\dfrac{x^2+x-2+4x-x^2}{x\left(x-2\right)}=\dfrac{-x^2+3x-2}{x\left(2-x\right)}:\dfrac{5x-2}{x\left(x-2\right)}=\dfrac{-\left(x-1\right)\left(x-2\right)}{x\left(2-x\right)}.\dfrac{x\left(x-2\right)}{5x-2}=\dfrac{\left(x-1\right)\left(x-2\right)}{x\left(x-2\right)}.\dfrac{x\left(x-2\right)}{5x-2}=\dfrac{\left(x-1\right)\left(x-2\right)}{5x-2}=\dfrac{x^2-3x+2}{5x-2}\)b)\(x^2-5x+6=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Với x = 2, ta có A = 0

Với x = 3, ta có A = 2/13