Question

d) |x2 – 1| + |x + 1| = 0.

Answer 1

d) Ta có: \(\left|x^2-1\right|+\left|x+1\right|=0\)

\(\Leftrightarrow\left|x-1\right|\cdot\left|x+1\right|+\left|x+1\right|=0\)

\(\Leftrightarrow\left|x+1\right|\cdot\left(\left|x-1\right|+1\right)=0\)

mà \(\left|x-1\right|+1>0\forall x\)

nên \(\left|x+1\right|=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

Vậy: x=-1

Answer 2

bn ơi đề là |x2-1|+|x+1|=0 hay là |x²-1|+|x+1|=0 vậy bn???

Answer 3

|2x-1| + |x+1| = 0\(\Leftrightarrow\left[{}\begin{matrix}\left|2x-1\right|=0.\left|x+1\right|\\\left|x+1\right|=0.\left|2x-1\right|\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|2x-1\right|=0\\\left|x+1\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1:2=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)

Vậy ......