Bài 2:
a: ĐKXĐ: x∉{0;-1;4}
Ta có: \(\frac{11}{x}=\frac{9}{x+1}+\frac{2}{x-4}\)
=>\(\frac{11}{x}=\frac{9\left(x-4\right)+2\left(x+1\right)}{\left(x+1\right)\left(x+4\right)}\)
=>\(\frac{11}{x}=\frac{9x-36+2x+2}{x^2+5x+4}\)
=>\(11\left(x^2+5x+4\right)=x\left(11x-34\right)\)
=>\(11x^2+55x+44=11x^2-34x\)
=>55x+44=-34x
=>89x=-44
=>\(x=-\frac{44}{89}\) (nhận)
b: ĐKXĐ: x<>4
\(\frac{14}{3x-12}-\frac{2+x}{x-4}=\frac{3}{8-2x}-\frac56\)
=>\(\frac{14}{3\left(x-4\right)}-\frac{x+2}{x-4}+\frac{3}{2\left(x-4\right)}=-\frac56\)
=>\(\frac{28}{6\left(x-4\right)}-\frac{6\left(x+2\right)}{6\left(x-4\right)}+\frac{9}{6\left(x-4\right)}=\frac{-5}{6}\)
=>\(\frac{37-6x-12}{6\left(x-4\right)}=-\frac56\)
=>\(\frac{25-6x}{6\left(x-4\right)}=\frac{-5}{6}\)
=>\(\frac{6x-25}{6x-24}=\frac56\)
=>6(6x-25)=5(6x-24)
=>36x-150=30x-120
=>6x=30
=>x=5(nhận)
Bài 1:
e: ĐKXĐ: x<>0; x<>-5
\(\frac{2x+5}{2x}-\frac{x}{x+5}=0\)
=>\(\frac{\left(2x+5\right)\left(x+5\right)-2x\cdot x}{2x\left(x+5\right)}=0\)
=>\(\left(2x+5\right)\left(x+5\right)-2x^2=0\)
=>\(2x^2+10x+5x+25-2x^2=0\)
=>15x=-25
=>\(x=-\frac53\) (nhận)
f: ĐKXĐ: x<>4/11
\(\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\)
=>\(\frac{12x+1}{11x-4}=\frac{20x+17}{18}-\frac{20x-8}{18}=\frac{25}{18}\)
=>25(11x-4)=18(12x+1)
=>275x-100=216x+18
=>59x=118
=>x=2(nhận)
