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\(A=\frac{2x^2+4x}{x^3-4x}+\frac{x^2-4}{x^2+2x}+\frac{2}{2-x}\left(x\ne0;x\ne\pm2\right)\)
\(A=\frac{2x^2+4x}{x\left(x^2-4\right)}+\frac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}-\frac{2}{x-2}\)
\(A=\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)^2\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}+\frac{x^3-2x^2-4x+8}{x\left(x-2\right)\left(x+2\right)}-\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{2x^2+4x+x^3-2x^2-4x+8-2x^2-4x}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{-2x^2-4x+8}{x\left(x-2\right)\left(x+2\right)}=\frac{-2x\left(x+2\right)+8}{x\left(x-2\right)\left(x+2\right)}=\frac{-2x+8}{x\left(x-2\right)}\)
Vậy \(A=\frac{-2x+8}{x\left(x-2\right)}\left(x\ne0;x\ne\pm2\right)\)
b) \(A=\frac{-2x+8}{x\left(x-2\right)}\left(x\ne0;x\ne\pm2\right)\)
Ta có: x=4 (tmđk) thay vào A ta có:
\(A=\frac{-2\cdot4+8}{4\left(4-2\right)}=\frac{-8+8}{4\cdot2}=\frac{0}{8}=0\)
Vậy A=0 với x=4
a: \(P=\left(\dfrac{3}{2\left(x+2\right)}-\dfrac{x}{x-2}+\dfrac{2x^2+3}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{4\left(x-2\right)}{2x-1}\)
\(=\left(\dfrac{3\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}-\dfrac{2x\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\dfrac{4x^2+6}{2\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{4\left(x-2\right)}{2x-1}\)
\(=\dfrac{3x-6-2x^2-4x+4x^2+6}{2\left(x+2\right)\left(x-2\right)}\cdot\dfrac{4\left(x-2\right)}{2x-1}\)
\(=\dfrac{2x^2-x}{x+2}\cdot\dfrac{2}{2x-1}=\dfrac{2x}{x+2}\)
b: Khi 4x2-1=0 thì (2x-1)(2x+1)=0
=>x=1/2(loại) và x=-1/2(nhận)
Khi x=-1/2 thì \(P=\left(2\cdot\dfrac{-1}{2}\right):\left(-\dfrac{1}{2}+2\right)=-1:\dfrac{3}{2}=-\dfrac{2}{3}\)
Với `x \ne +-2,x \ne 1/2,x \ne0`. Ta có:
`(3/[2x+4]+x/[2-x]+[2x^2+3]/[x^2-4]):[2x-1]/[4x-8]`
`=(3/[2(x+2)]-x/[x-2]+[2x^2+3]/[(x-2)(x+2)]).[4(x-2)]/[2x-1]`
`=[3(x-2)-2x(x+2)+2(2x^2+3)]/[x(x-2)(x+2)].[4(x-2)]/[2x-1]`
`=[3x-6-2x^2-4x+4x^2+6]/[x(x+2)]. 4/[2x-1]`
`=[2x^2-x]/[x(x+2)]. 4/[2x-1]`
`=[x(2x-1)]/[x(x+2)] . 4/[2x-1]`
`=4/[x+2]`