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

a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)

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

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

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

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

Có vài bước mình làm tắc á nha :>

a)

A=\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5x-5}\)

\(\Leftrightarrow\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5\left(x-1\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+1\\x=0-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

MTC: 5(x-1)(x+1)

\([\dfrac{5\left(x+1\right)\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}-\dfrac{5\left(x-1\right)\left(x-1\right)}{5\left(x-1\right)\left(x+1\right)}]\div\dfrac{2x\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow[5\left(x+1\right)\left(x+1\right)-5\left(x-1\right)\left(x-1\right)]\div2x\left(x+1\right)\)

\(\Leftrightarrow[5\left(x+1\right)^2-5\left(x-1\right)^2]\div2x^2+2x\)

\(\Leftrightarrow[5\left(x^2+2x+1\right)-5\left(x^2-2x+1\right)]\div2x^2+2x\)

\(\Leftrightarrow(5x^2+10x+5-5x^2+10x-5)\div2x^2+2x\)

\(\Leftrightarrow20x\div\left(2x^2+2x\right)\)

\(\Leftrightarrow10x+10\)

a: ĐKXĐ: x∉{0;2;-2;3}

b: \(A=\left(\frac{4x}{x+2}+\frac{8x^2}{4-x^2}\right):\left(\frac{x-1}{x^2-2x}-\frac{2}{x}\right)\)

\(=\frac{4x\left(x-2\right)-8x^2}{\left(x-2\right)\left(x+2\right)}:\left(\frac{x-1}{x\left(x-2\right)}-\frac{2}{x}\right)\)

\(=\frac{4x^2-8x-8x^2}{\left(x-2\right)\left(x+2\right)}:\frac{x-1-2\left(x-2\right)}{x\left(x-2\right)}\)

\(=\frac{-4x^2-8x}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x\left(x-2\right)}{x-1-2x+4}=\frac{-4x\left(x+2\right)}{\left(x+2\right)}\cdot\frac{x}{-x+3}=\frac{4x^2}{x-3}\)

c: \(x^2+2x=15\)

=>\(x^2+2x-15=0\)

=>(x+5)(x-3)=0

=>x=-5(nhận) hoặc x=3(loại)

Khi x=-5 thì \(A=4\cdot\frac{\left(-5\right)^2}{-5-3}=4\cdot\frac{25}{-8}=\frac{-100}{8}=-12,5\)

d: |A|>A

=>A<0

=>\(\frac{4x^2}{x-3}<0\)

=>x-3<0

=>x<3

Kết hợp ĐKXĐ, ta được: x<3 và x∉{0;2;-2}

a) Ta có: \(P=\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\cdot\dfrac{1-x^2}{2}\)

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

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

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

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

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

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

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

\(=\dfrac{-x^3+5x-6}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-\left(x^3-5x+6\right)}{2\left(x-1\right)\left(x+1\right)}\)

a: \(M=12-x^3+6x-x^2-4x+x^3+2x^2\)

\(=\left(-x^3+x^3\right)+\left(2x^2-x^2\right)+6x-4x+12\)

\(=x^2+2x+12\)

b: \(M=x^2+2x+12=x^2+2x+1+11=\left(x+1\right)^2+11\ge11\forall x\)

Dấu '=' xảy ra khi x+1=0

=>x=-1

=>M0=11; x0=-1

b+2Mo-9xo

=2+2*11-9*(-1)

=24+9

=33