Question

Answer 1

`a)` Với `x \ne 2` có:

`[x-1]/[x^2+2x+4]=[(x-1)(x-2)]/[(x-2)(x^2+2x+4)]=[x^2-2x-x+2]/[x^3-8]`

                                    `=[x^2-3x+2]/[x^3-8]`

  `=>A=x^2-3x+2`

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`b)` Với `x \ne 1/2;x \ne 1;x \ne 3` có:

`[2x-1]/[x-3].1/B=1/[x^2-4x+3]`

`1/B=1/[(x-3)(x-1)]:[2x-1]/[x-3]`

`1/B=1/[(x-1)(2x-1)]=1/[2x^2-x-2x+1]=1/[2x^2-3x+1]`

    `=>B=2x^2-3x+1`