\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{\left(x-1\right)\left(x+1\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
Ta có : \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{16}{\left(x-1\right)\left(x+1\right)}\)
`=> x^2 + 2x +1-(x^2 -2x+1)=16`
`<=> x^2 +2x+1-x^2 +2x-1=16`
`<=>4x =16`
`<=>x=4(tm)`
Vậy phương trình có nghiệm `x=4`
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