Đáp án:
Giải thích các bước giải:
`b)ĐKXĐ:x\ne 1;x\ne 2;x\ne 3`
`\frac{3}{(x-1)(x-2)}+\frac{2}{(x-3)(x-1)}=\frac{1}{(x-1)(x-2)}`
`\Rightarrow 3(x-3)+2(x-2)=x-3`
`\Leftrightarrow 3x-9+2x-4=x-3`
`\Leftrightarrow 5x-13=x-3`
`\Leftrightarrow 4x=10`
`\Leftrightarrow x=10/4=5/2(tm)`
Vậy `S=\{5/2\}`
`d)ĐKXĐ:x\ne 3;x\ne -3;x\ne -7/2`
`\frac{13}{(x-3)(2x+7)}+\frac{1}{2x+7}=\frac{6}{(x-3)(x+3)}`
`\Rightarrow 13(x+3)+(x-3)(x+3)=6(2x+7)`
`\Leftrightarrow 13x+39+x^2-9=12x+42`
`\Leftrightarrowx^2+13x+30=12x+42`
`\Leftrightarrow x^2+x-12=0`
`\Leftrightarrow (x+4)(x-3)=0`
`\Leftrightarrow x+4=0`hoặc `x-3=0`
`\Leftrightarrow x=-4(tm)` hoặc `x=3(ktm)`
Vậy `S=\{-4\}`
