Question

tìm x

x-1/x+1=0

Answer 1

\(\frac{x-1}{x+1}=0\\ \Rightarrow x-1=0\\ x=0+1\\ \Rightarrow x=1\\ @@\)

Answer 2

⇔ \(\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)

⇔ \(\frac{x^2-x-x+1}{x^2-1^2}=0\)

⇔ \(\frac{x^2-2x+1}{x^2-1^2}\) = 0

⇔ \(\frac{x^2-2x.1+1^2}{x^2-1^2}\) = 0

⇔\(\frac{\left(x-1\right)^2}{x^2-1^2}=0\)

⇔ 0 = 0

KO BT ĐÚNG KO NHÉ

Answer 3

\(\frac{x-1}{x+1}=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x\ne-1\end{matrix}\right.\)
Vậy \(x=1vàx\ne-1\) thì phân thức \(\frac{x-1}{x+1}=0\)