`a)[x+1]/[x-1]-[x-1]/[x+1]=16/[x^2-1]` `ĐK: x \ne +-1`
`=>(x+1)^2-(x-1)^2=16`
`<=>(x+1-x+1)(x+1+x-1)=16`
`<=>2.2x=16`
`<=>x=4` (t/m)
`b)1/[x-1]-[3x^2]/[x^3-1]=[2x]/[x^2+x+1]` `ĐK: x \ne 1`
`=>x^2+x+1-3x^2=2x(x-1)`
`<=>x^2+x+1-3x^2=2x^2-2x`
`<=>4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>(x-1)(4x+1)=0`
`<=>[(x=1(ko t//m)),(x=-1/4(t//m)):}`
a)
\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)
\(< =>\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{16}{\left(x-1\right)\left(x+1\right)}\)
\(< =>x^2+2x+1-x^2+2x-1=16\)
\(< =>4x=16\\ < =>x=4\)
b)
\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\) (đk x khác 1)
\(< =>\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(< =>x^2+x+1-3x^2=2x^2-2x\)
\(< =>x^2+x+1-3x^2-2x^2+2x=0\)
\(< =>-4x^2+3x+1=0\)
\(< =>-\left(4x^2-3x-1\right)=0\)
\(< =>-\left(4x^2-4x+x-1\right)=0\\ < =>-\left[4x\left(x-1\right)+\left(x-1\right)\right]=0\\ < =>-\left(x-1\right)\left(4x+1\right)=0\\ < =>\left[{}\begin{matrix}x-1=0< =>x=1\left(loai\right)\\4x+1=0< =>x=-\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)
