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\(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\cdot\dfrac{2-x}{x}\)
\(=\dfrac{x+2+2x+x-2}{-\left(2-x\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{-\left(x+2\right)\cdot x}=\dfrac{-4}{x+2}\)
a: \(P=\dfrac{4x-6-x+1}{2x-3}:\left(\dfrac{6x+1}{2x^2-3x+2x-3}+\dfrac{x}{x+1}\right)\)
\(=\dfrac{3x-5}{2x-3}:\left(\dfrac{6x+1}{\left(x+1\right)\left(2x-3\right)}+\dfrac{x}{x+1}\right)\)
\(=\dfrac{3x-5}{2x-3}:\dfrac{6x+1+2x^2-3x}{\left(x+1\right)\left(2x-3\right)}\)
\(=\dfrac{3x-5}{\left(2x-3\right)}\cdot\dfrac{\left(2x-3\right)\left(x+1\right)}{2x^2+3x+1}\)
\(=\dfrac{3x-5}{2x+1}\)
b: \(P-\dfrac{3}{2}=\dfrac{3x-5}{2x+1}-\dfrac{3}{2}=\dfrac{6x-10-6x-3}{2\left(2x+1\right)}=\dfrac{-7}{2\left(2x+1\right)}\)
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right):\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\left(x\ne-1;x\ne0;x\ne-2\right)\)
\(=\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right):\frac{3x^3-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\left(\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3x+3}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)\(:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2-x+1-3+3x+3}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\frac{\left(x+1\right)\left(x+2\right)}{3\left(x^2-x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+2\right)^2\left(x+1\right)}{3\left(x^2-x+1\right)^2}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(\left(\frac{2x^2+1}{x^2-1}-\frac{1}{x-1}\right):\left(1-\frac{x^2+4}{x^2+x+1}\right)\)
\(=\left[\frac{2x^2+1}{\left(x-1\right)\left(x+1\right)}-\frac{x+1}{\left(x-1\right)\left(x+1\right)}\right]:\frac{x^2+x+1-x^2-4}{x^2+x+1}\)
\(=\frac{2x^2+1-x-1}{\left(x-1\right)\left(x+1\right)}:\frac{x-3}{x^2+x+1}\)
\(=\frac{2x^2-x}{\left(x-1\right)\left(x+1\right)}.\frac{x^2+x+1}{x-3}\)
bài này đến đây cậu làm tiếp chư tôi ko tách ra đc nữa