8: ĐKXĐ: x<>2
\(\dfrac{x^2+6x-16}{x-2}=x+8\)
=>\(\dfrac{x^2+8x-2x-16}{x-2}=x+8\)
=>\(\dfrac{\left(x+8\right)\left(x-2\right)}{x-2}=x+8\)
=>x+8=x+8
=>0x=0(luôn đúng)
Vậy: \(x\in R\backslash\left\{2\right\}\)
9: ĐKXĐ: x<>2
\(\dfrac{x-1}{x-2}-3+x=\dfrac{1}{x-2}\)
=>\(\dfrac{x-1}{x-2}+\dfrac{\left(x-3\right)\left(x-2\right)}{x-2}=\dfrac{1}{x-2}\)
=>\(x-1+x^2-5x+6=1\)
=>\(x^2-4x+4=0\)
=>\(\left(x-2\right)^2=0\)
=>x-2=0
=>x=2(loại)
10: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)
=>\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{\left(x-1\right)\left(x+1\right)}\)
=>\(\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{16}{\left(x-1\right)\left(x+1\right)}\)
=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)
=>\(x^2+2x+1-x^2+2x-1=16\)
=>4x=16
=>x=4(nhận)
