Bài 4: Giá trị tuyệt đối của một số hữu tỉ. Cộng, trừ, nhân, chia số thập phân

CT

Tìm x:

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

b) \(\left|x-\dfrac{1}{2}\right|+\dfrac{5}{6}=1\dfrac{1}{2}\)

c) \(\left|x-\dfrac{1}{2}\right|+\dfrac{4}{5}=\left|\left(-3,2\right)+\dfrac{2}{5}\right|\)

NH
2 tháng 9 2017 lúc 12:51

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-0,7\\x=-1,7\end{matrix}\right.\)

Vậy ....

b, \(\left|x-\dfrac{1}{2}\right|+\dfrac{5}{6}=1\dfrac{1}{2}\)

\(\Leftrightarrow\left|x-\dfrac{1}{2}\right|=1\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Leftrightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{2}{3}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{-1}{6}\end{matrix}\right.\)

Vậy .....

c, \(\left|x-\dfrac{1}{2}\right|+\dfrac{4}{5}=\left|-3,2+\dfrac{2}{5}\right|\)

\(\left|x-\dfrac{1}{2}\right|+\dfrac{4}{5}=\left|-2,8\right|\)

\(\left|x-\dfrac{1}{2}\right|+\dfrac{4}{5}=2,8\)

\(\left|x-\dfrac{1}{2}\right|=2,8-\dfrac{4}{5}\)

\(\left|x-\dfrac{1}{2}\right|=2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-8}{5}\end{matrix}\right.\)

Vậy ...

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