Đáp án:
$a)$ $S=$ {$3$}
$b)$ $S=$ {$2,±3$}
$c)$ $S=$ {$\frac{-5}{3}$$,-1$}
$d)$ $S=$ {$\frac{-2}{5}$}
Giải thích các bước giải:
$a)$ $4x-1=2x+5$
⇔ $4x-2x=5+1$
⇔ $2x=6$
⇔ $x=3$
$KL:$ $S=$ {$3$}
$b)$ $x^2(x-2)=9x-18$
⇔ $x^2(x-2)-9x+18=0$
⇔ $x^2(x-2)-9(x-2)=0$
⇔ $(x-2)(x^2-9)=0$
⇔ $(x-2)(x-3)(x+3)=0$
⇔\(\left[ \begin{array}{l}x=2\\x=3\\x=-3\end{array} \right.\)
$KL:$ $S=$ {$2,±3$}
$c)$ $|2x+3|=x+2$
$TH1:$ $2x+3=x+2$ $($$x$ $\geq$ $\frac{-3}{2}$$)$
⇔ $2x-x=2-3$
⇔ $x=-1$$(tm)$
$TH2:$ $-2x-3=x+2$ $($$x<$ $\frac{-3}{2}$$)$
⇔ $-2x-x=2+3$
⇔ $-3x=5$
⇔ $x=$ $\frac{-5}{3}$$(tm)$
$KL:$ $S=$ {$\frac{-5}{3}$$,-1$}
$d)$ $\frac{x-5}{x-1}$ $=$ $\frac{2x-3}{x^2-1}$ $+$ $\frac{x}{x+1}$ $($$ĐK:$ $x$ $\neq$ $±1$$)$
⇔$\frac{x-5}{x-1}$ $-$ $\frac{2x-3}{x^2-1}$ $-$ $\frac{x}{x+1}$ $=0$
⇔ $\frac{(x-5)(x+1)-2x+3-x(x-1)}{(x-1)(x+1)}$ $=0$
⇔ $x^2-5x+x-5-2x+3-x^2+x=0$ <=> $-5x-2=0$
⇔ $-5x=2$
⇔ $x=$ $\frac{-2}{5}$$(tmđk)$
$KL:$ $S=$ {$\frac{-2}{5}$}
