Question

[lê tiến trường]

a)\(|2.5-x|=1.3\)

b)1.6-\(|x-0.2|=0\)

c)\(|x-1.5|+|2.5-x|=0\)

Answer 1

[Nguyễn Thanh Hằng]

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=3,8\end{matrix}\right.\)

Vậy .

b/ \(1,6-\left|x-0,2\right|=0\)

\(\Leftrightarrow\left|x-0,2\right|=1,6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0,2=1,6\\x-0,2=-1,6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,8\\x=-1,4\end{matrix}\right.\)

Vậy ....

c/ \(\left|x-1,5\right|+\left|2,5-x\right|=0\)

Mà \(\left\{{}\begin{matrix}\left|x-1,5\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-1,5\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\) (loại)

Vậy ..........

Answer 2

[Hải Đăng]

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

\(\Rightarrow\left[{}\begin{matrix}2,5-x=1,3\\2,5-x=-1,3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,2\\x=3,8\end{matrix}\right.\)

Vậy .....

\(b)1,6-\left|x-0,2\right|=0\)

\(\Rightarrow\left|x-0,2\right|=1,6-0\)

\(\Rightarrow\left|x-0,2\right|=0\)

\(\Rightarrow x-0,2=0\)

\(\Rightarrow x=0,2\)

Vậy .......

\(c)\left|x-1,5\right|+\left|2,5-x\right|=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)

Vậy .....

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