Bài 6:
a: ĐKXĐ: x∉{1;-1}
Ta có: \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\)
=>\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{\left(x-1\right)\left(x+1\right)}\)
=>\(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=\frac{16}{\left(x+1\right)\left(x-1\right)}\)
=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)
=>(x+1+x-1)(x+1-x+1)=16
=>\(2\cdot2x=16\)
=>4x=16
=>x=4(nhận)
b: ĐKXĐ: x∉{0;-1}
ta có: \(\frac{2x-4}{x\left(x+1\right)}+\frac{4}{x}=\frac{8-x}{x+1}\)
=>\(\frac{2x-4}{x\left(x+1\right)}+\frac{4\left(x+1\right)}{x\left(x+1\right)}=\frac{x\left(8-x\right)}{x\left(x+1\right)}\)
=>\(2x-4+4\left(x+1\right)=x\left(8-x\right)\)
=>\(2x-4+4x+4=8x-x^2\)
=>\(6x-8x+x^2=0\)
=>\(x^2-2x=0\)
=>x(x-2)=0
=>x=0(loại) hoặc x=2(nhận)
c: ĐKXĐ: x∉{0;3}
Ta có: \(\frac{2x+1}{x}-\frac{x-2}{x-3}=\frac{2x-3}{x^2-3x}\)
=>\(\frac{\left(2x+1\right)\left(x-3\right)-x\left(x-2\right)}{x\left(x-3\right)}=\frac{2x-3}{x\left(x-3\right)}\)
=>\(\left(2x+1\right)\left(x-3\right)-x\left(x-2\right)=2x-3\)
=>\(2x^2-6x+x-3-x^2+2x-2x+3=0\)
=>\(x^2-5x=0\)
=>x(x-5)=0
=>x=0(loại) hoặc x=5(nhận)