a: ĐKXĐ: x∉{0;-1;4}
Ta có: \(\frac{11}{x}=\frac{9}{x+1}+\frac{2}{x-4}\)
=>\(\frac{11}{x}=\frac{9\left(x-4\right)+2\left(x+1\right)}{\left(x+1\right)\left(x-4\right)}\)
=>\(\frac{11}{x}=\frac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)
=>\(11\left(x+1\right)\left(x-4\right)=x\left(11x-34\right)\)
=>\(11\left(x^2-3x-4\right)=11x^2-34x\)
=>\(11x^2-33x-44=11x^2-34x\)
=>-33x+34x=44
=>x=44(nhận)
b: ĐKXĐ: x∉{1/3;-1/3}
Ta có: \(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
=>\(\frac{-12}{\left(3x-1\right)\left(3x+1\right)}=-\frac{3x-1}{3x+1}+\frac{3x+1}{3x-1}\)
=>\(\frac{-12}{\left(3x-1\right)\left(3x+1\right)}=\frac{-\left(3x-1\right)^2+\left(3x+1\right)^2}{\left(3x-1\right)\left(3x+1\right)}\)
=>\(\left(3x+1\right)^2-\left(3x-1\right)^2=12\)
=>\(9x^2+6x+1-9x^2+6x-1=12\)
=>12x=12
=>x=1(nhận)
c: ĐKXĐ: x∉{1;-1}
Ta có: \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\)
=>\(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=\frac{16}{\left(x-1\right)\left(x+1\right)}\)
=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)
=>\(x^2+2x+1-x^2+2x-1=16\)
=>4x=16
=>x=4(nhận)
