`B12:`
`a)` Với `x \ne +-1` có:
`A=4/[x+1]+2/[1-x]-[x-5]/[x^2-1]`
`A=[4(x-1)-2(x+1)-x+5]/[(x-1)(x+1)]`
`A=[4x-4-2x-2-x+5]/[(x-1)(x+1)]`
`A=[x-1]/[(x-1)(x+1)]=1/[x+1]`
~~~~~~~~~~~~~~~~~~~~~~~
`b)` Với `x \ne +-1` có:
`A^2=2A<=>(1/[x+1])^2=2. 1/[x+1]<=>1/[(x+1)^2]-2/[x+1]=0`
`<=>[1-2(x+1)]/[(x+1)^2]=0`
`=>1-2x-2=0<=>x=-1/2` (t/m)
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`B13:`
`a)` Với `x \ne 1,x \ne -3` có:
`M=[3x^2+5x-4]/[(x+3)(x-1)]-[x+1]/[x+3]-[x+3]/[x-1]`
`M=[3x^2+5x-4-(x+1)(x-1)-(x+3)^2]/[(x+3)(x-1)]`
`M=[3x^2+5x-4-x^2+1-x^2-6x-9]/[(x+3)(x-1)]`
`M=[x^2-x-12]/[(x+3)(x-1)]`
`M=[(x+3)(x-4)]/[(x+3)(x-1)]=[x-4]/[x-1]`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`b)` Với `x \ne 1,x \ne -3` có:
`M=0,5<=>[x-4]/[x-1]=1/2<=>[x-4]/[x-1]-1/2=0`
`<=>[2(x-4)-x+1]/[2(x-1)]=0`
`=>2x-8-x+1=0<=>x=7` (t/m)
