`a)|x|=2<=>x=+-2` mà `x \ne -2`
`=>` Thay `x=2` vào `B` có: `B=[2.2+2]/[2+2]=3/2`
`b)` Với `x \ne +-1` có:
`A=[x+1]/[2(x-1)]-1/[2(x^2-1)]`
`A=[(x+1)^2-1]/[2(x-1)(x+1)]`
`A=[x^2+2x]/[2(x-1)(x+1)]`
`c)` Với `x \ne +-1;x \ne -2` có:
`P=A.B=[x^2+2x]/[2(x-1)(x+1)].[2x+2]/[x+2]`
`=[x(x+2)]/[2(x-1)(x+1)].[2(x+1)]/[x+2]`
`=x/[x-1]`
`|P|=3<=>|x/[x-1]=3`
`<=>[(x/[x-1]=3),(x/[x-1]=-3):}`
`<=>[(x=3x-3),(x=-3x+3):}<=>{(x=3/2),(x=3/4):}` (t/m)
a)
có
\(\left|x\right|=2\\ =>\left[{}\begin{matrix}x=2\left(tm\right)\\x=-2\left(loai\right)\end{matrix}\right.\)
thay x=2 vào biểu thức B ta có
\(\dfrac{2\cdot2+2}{2+2}=\dfrac{6}{4}=1,5\)
b)
\(\dfrac{x+1}{2x-2}+\dfrac{1}{2-2x^2}\\ =\dfrac{x+1}{2\left(x-1\right)}+\dfrac{1}{2\left(1-x\right)\left(1+x\right)}\)
\(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}-\dfrac{1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x+1-1}{2\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x\left(x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)
c)
\(P=\dfrac{x\left(x+2\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2x+2}{x+2}\)
\(=\dfrac{x\left(x+2\right)\cdot2\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)\cdot\left(x+2\right)}\\ =\dfrac{x}{x-1}\)
\(\left|P\right|=3\\ =>\left[{}\begin{matrix}P=-3\left(tm\right)\\P=3\left(tm\right)\end{matrix}\right.\)
với P=-3 ta có
\(-3=\dfrac{x}{x-1}\\ =>-3\cdot\left(x-1\right)=x\)
\(=>-3x+3-x=0\)
\(=>-4x+3=0\\ =>x=\dfrac{3}{4}\)
với P=3 ta có
\(3=\dfrac{x}{x-1}\\ =>3\cdot\left(x-1\right)=x\)
\(=>3x-3-x=0\)
\(=>2x-3=0\\ =>x=\dfrac{3}{2}\)
