\(\dfrac{1}{x+2}+\dfrac{2}{2-x}=\dfrac{2x-3}{x^2-4}\)
\(\left(-\right)\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-3}{x^2-4}\)
\(\left(-\right)\dfrac{x-2}{x^2-4}-\dfrac{2x+4}{x^2-4}-\dfrac{2x-3}{x^2-4}=0\)
\(\left(-\right)x-2-2x-4-2x+3=0\)
\(\left(-\right)-3x-3=0\)
\(\left(-\right)-3x=3\)
\(\left(-\right)x=-1\)
Đặt A = \(\dfrac{1}{x+1}+\dfrac{2}{2-x}=\dfrac{2x-3}{x^2-4}\)
ĐKXĐ: x#2, x#-2.
<=> \(\dfrac{1}{x+1}-\dfrac{2}{x-2}=\dfrac{2x-3}{\left(x+2\right)\left(x-2\right)}\)
<=> \(\dfrac{x-2-2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x-3}{\left(x+2\right)\left(x-2\right)}\)
<=> -x - 6 = 2x - 3
<=> -3x = 3
<=> x = -1 (T/m ĐKXĐ)
Vậy S = {-1}.
Mk giải bài này có j sai thì bảo mk ha!!!
