Question

Tìm x:

a. \(|\dfrac{5}{3}x|=|\dfrac{1}{6}|\)

b.\(|\dfrac{3}{4}x-\dfrac{3}{4}|-\dfrac{3}{4}=|-\dfrac{3}{4}|\)

c.\(|x+\dfrac{3}{5}|-|x-\dfrac{7}{3}|=0\)

Answer 1

Answer 2

a) \(\left|\dfrac{5}{3}x\right|=\left|\dfrac{1}{6}\right|\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=\dfrac{-1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|-\dfrac{3}{4}\right|\)

\(\Rightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)

\(\Rightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{2}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\\\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=-\dfrac{3}{4}\\\dfrac{3}{4}x=\dfrac{9}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

c) \(\left|x+\dfrac{3}{5}\right|=\left|x-\dfrac{7}{3}\right|\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{5}=x-\dfrac{7}{3}\\x+\dfrac{3}{5}=-x+\dfrac{7}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=\dfrac{13}{15}\end{matrix}\right.\)

Answer 3

a. \(\left|\dfrac{5}{3}x\right|=\left|\dfrac{1}{6}\right|\)

\(\Rightarrow\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{-1}{6}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{10}\\x=\dfrac{-1}{15}\end{matrix}\right.\)

Vậy...

b. \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|-\dfrac{3}{4}\right|\)

\(\Rightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)

\(\Rightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}\)

\(\Rightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{2}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\\\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{4}\\\dfrac{3}{4}x=\dfrac{-3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy...

c. \(\left|x+\dfrac{3}{5}\right|-\left|x-\dfrac{7}{3}\right|=0\)

Để \(\left|x+\dfrac{3}{5}\right|-\left|x-\dfrac{7}{3}\right|=0\) thì

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{5}\right|=0\\\left|x-\dfrac{7}{5}\right|=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{5}=0\\x-\dfrac{7}{5}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{5}\\x=\dfrac{7}{5}\end{matrix}\right.\) ( vô lí )

Vậy không có giá trị nào thỏa mãn.