Bài 12:
\(\Leftrightarrow\dfrac{1}{\left(2x+7\right)\left(x-1\right)}-\dfrac{2}{\left(x-1\right)\left(x+1\right)}=\dfrac{3}{\left(2x-7\right)\left(x+1\right)}\)
\(\Leftrightarrow\left(2x-7\right)\left(x+1\right)-2\left(4x^2-49\right)=3\left(2x+7\right)\left(x-1\right)\)
\(\Leftrightarrow2x^2+2x-7x-7-8x^2+98=3\left(2x^2-2x+7x-7\right)\)
\(\Leftrightarrow-6x^2-5x+91=6x^2+15x-21\)
\(\Leftrightarrow-12x^2-20x+112=0\)
\(\Leftrightarrow3x^2+5x-28=0\)
\(\Leftrightarrow3x^2+12x-7x-28=0\)
=>(x+4)(3x-7)=0
=>x=-4 hoặc x=7/3
\(\dfrac{1}{2x^2+5x-7}-\dfrac{2}{x^2-1}=\dfrac{3}{2x^2-5x-7}\)
\(đk:x\ne\pm1;x\ne-\dfrac{7}{2};x\ne\dfrac{7}{2}\)
\(pt\Leftrightarrow\dfrac{1}{\left(x-1\right)\left(2x+7\right)}-\dfrac{2}{\left(x-1\right)\left(x+1\right)}=\dfrac{3}{\left(x+1\right)\left(2x-7\right)}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(2x-7\right)}{\left(x-1\right)\left(x+1\right)\left(2x+7\right)\left(2x-7\right)}-\dfrac{2\left(2x-7\right)\left(2x+7\right)}{\left(x-1\right)\left(x+1\right)\left(2x-7\right)\left(2x+7\right)}=\dfrac{3\left(x-1\right)\left(2x+7\right)}{\left(x+1\right)\left(x-1\right)\left(2x-7\right)\left(2x+7\right)}\Rightarrow\left(x+1\right)\left(2x-7\right)-2\left(2x-7\right)\left(2x+7\right)=3\left(x-1\right)\left(2x+7\right)\Leftrightarrow x=.....\)
\(B13\left(tương-tự\right)\)
Bài 13:
ĐKXĐ: $x\neq 1; -5; 9$
PT \(\Leftrightarrow \frac{2x-1}{(x-1)(x+5)}+\frac{x-2}{(x-9)(x-1)}=\frac{3x-12}{(x-9)(x+5)}\)
\(\Leftrightarrow \frac{(2x-1)(x-9)}{(x-1)(x+5)(x-9)}+\frac{(x-2)(x+5)}{(x-9)(x-1)(x+5)}=\frac{(3x-12)(x-1)}{(x-9)(x+5)(x-1)}\)
\(\Rightarrow (2x-1)(x-9)+(x-2)(x+5)=(3x-12)(x-1)\)
$\Leftrightarrow -x-13=0$
$\Leftrightarrow x=-13$ (tm)
