Question

[pham thi ngoc]

a) \(\left|3\cdot x-1\right|=\left|5-2\cdot x\right|\)

b) \(\left|x+2\right|-\left|x+7\right|=0\)

c) \(\left|2\cdot x-1\right|+x=2\)

d) \(\left|x+1\right|+\left|2\cdot y-1\right|=0\)

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Answer 1

[Nguyễn Thanh Hằng]

a/ \(\left|3x-1\right|=\left|5-2x\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5-2x\\3x-1=-5+2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2x=5+1\\3x-2x=-5+1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}5x=6\\x=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-4\end{matrix}\right.\)

Vậy ......

b/ \(\left|x+2\right|-\left|x+7\right|=0\)

\(\Leftrightarrow\left|x+2\right|=\left|x+7\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=x+7\\x+2=-x-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-x=7-2\\x+x=-7-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\2x=-9\end{matrix}\right.\)

\(\Leftrightarrow x=-\dfrac{9}{2}\)

Vậy ...............

c/ \(\left|2x-1\right|+x=2\)

\(\Leftrightarrow\left|2x-1\right|=2-x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x=3\\x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

Vậy ..