`a)` Với `x \ne +-3, x \ne -1` có:
`A=(x/[x+3]-2/[x-3]+[x^2-1]/[9-x^2]):(2-[x+5]/[3+x])`
`A=[x(3-x)+2(3+x)+x^2-1]/[(3-x)(3+x)]:[2(3+x)-x-5]/[3+x]`
`A=[3x-x^2+6+2x+x^2-1]/[(3-x)(3+x)]:[6+2x-x-5]/[3+x]`
`A=[5x+5]/[(3-x)(3+x)].[3+x]/[x+1]`
`A=[5(x+1)]/[(3-x)(x+1)]=5/[3-x]`
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`b)x^2-x-2=0`
`<=>x^2-2x+x-2=0`
`<=>(x-2)(x+1)=0`
`<=>`$\left[\begin{matrix} x=2(t/m)\\ x=-1(ko t/m)\end{matrix}\right.$
`=>` Thay `x=2` vào `A` có: `A=5/[3-2]=5`
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`c)` Với `x \ne +-3;x \ne -1` có:
`A=1/2<=>5/[3-x]=1/2<=>3-x=10<=>x=-7` (t/m)
