Đáp án:
Giải thích các bước giải:
$\text{b, $\frac{4x-2}{3}$ - x+1= $\frac{1-5x}{4}$ }$
$\text{⇔$\frac{16x-8}{12}$ - $\frac{12x}{12}$ + $\frac{12}{12}$ = $\frac{3-15x}{12}$ }$
$\text{⇔$\frac{16x-8-12x+12}{12}$ = $\frac{3-15x}{12}$ }$
$\text{⇔4x+4=3-15x}$
$\text{⇔19x=-1}$
$\text{⇔x= $\frac{-1}{19}$ }$
$\text{Vậy S= {$\frac{-1}{19}$} }$
$\text{d, $\frac{x-1}{x+3}$ - $\frac{x}{x-3}$ = $\frac{7x-3}{9-x²}$ (ĐKXĐ: x $\neq$ ±3)}$
$\text{⇔$\frac{x-1}{x+3}$ + $\frac{x}{3-x}$ = $\frac{7x-3}{(3-x)(3+x)}$ }$
$\text{⇔$\frac{(x-1)(3-x)}{(x+3)(3-x)}$ + $\frac{x(x+3)}{(x+3)(3-x)}$ = $\frac{7x-3}{(x+3)(3-x)}$ }$
$\text{⇒3x-x²-3+x+x²+3x=7x-3}$
$\text{⇔7x-3=7x-3(luôn đúng)}$
$\text{⇔x ∈ R}$
$\text{Vậy S ∈ R}$
e, I 2x-1 I=5
⇔\(\left[ \begin{array}{l}2x-1=5\\2x-1=-5\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=-2\end{array} \right.\)
$\text{Vậy S= {3; -2}}$
h, I 5x-4 I= 4-5x
⇔\(\left[ \begin{array}{l}5x-4=4-5x\\5x-4=-4+5\end{array} \right.\)
⇔\(\left[ \begin{array}{l}10x=8\\5x-4=5x-4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{4}{5} \\0=0(luôn đúng)\end{array} \right.\)
$\text{Vậy S={$\frac{4}{5}$ }}$
