a, \(|\frac{x+1}{x-1}|=1\)
\(\Rightarrow|\frac{x+1}{x-1}|=1,Đkxđ:x\ne1\)
\(\Rightarrow\left[{}\begin{matrix}\frac{x+1}{x-1}=1\\\frac{x+1}{x-1}=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=0\end{matrix}\right.,x\ne1\)
b,mk k hiểu
c,\(2|x|-|y-3|=3\)
Đặt y =0
\(\Rightarrow2|x|-|0-3|=3\)
\(\Rightarrow2|x|-|-3|=3\)
\(\Rightarrow2|x|-3=3\)
\(\Rightarrow2|x|=6\)
\(\Rightarrow|x|=3\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
d, \(|x+1|+|x-3|=x\)
\(\Rightarrow|x+1|+|x-3|-x=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1+x-3-x=0,x+1\ge0;x-3\ge0\\-\left(x+1\right)+x-3-x=0;x+1< 0;x-3\ge0\\x+1-\left(x-3\right)-x=0;x+1\ge0,x-3< 0\\-\left(x+1\right)-\left(x-3\right)-x=0;x+1< 0,x-3< 0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\\x=4\\x=\frac{2}{3}\end{matrix}\right.\)
Ko thỏa mãn
\(\Rightarrow x\in\varnothing\)
a) \(\left|\frac{x+1}{x-1}\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x+1}{x-1}=1\\\frac{x+1}{x-1}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=x-1\\x+1=-x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0=0\\2x=0\end{matrix}\right.\)
\(\Leftrightarrow x=0\)
Vậy : \(x=0\)
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