Đáp án + Giải thích các bước giải:
`b)` `frac{1-6x}{x-2}+frac{9x+4}{x+2}=frac{x(2-3x)-1}{4-x^2}` ĐKXĐ: `x\ne±2`
`<=>frac{(1-6x)(x+2)}{(x-2)(x+2)}+frac{(9x+4)(x-2)}{(x-2)(x+2)}=frac{-x(2-3x)-1}{(x-2)(x+2)}`
`=>(1-6x)(x+2)+(9x+4)(x-2)=-x(2-3x)+1`
`<=>x+2-6x^2-12x+(9x^2-18x+4x-8)=-2x+3x^2+1`
`<=>-6x^2-11x+2+9x^2-14x-8=3x^2-2x+1`
`<=>3x^2-25x-6=3x^2-2x+1`
`<=>3x^2-25x-6-3x^2+2x-1=0`
`<=>-23x-7=0`
`<=>-23x=7`
`<=>x=-7/23` `text{( TMĐK )}`
Vậy `S={-7/23}`
`c)` `|x+2|=2x-10` ĐKXĐ: `x\geq5`
`<=>`\(\left[ \begin{array}{l}x+2=2x-10\\x+2=-2x+10\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x-2x=-10-2\\x+2x=10-2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}-x=-12\\3x=8\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=12(\text{nhận)}\\x=\dfrac{8}{3}(\text{ loại )}\end{array} \right.\)
Vậy `S={12}`
