Question

Tìm x \(\in\) , biết |\(\dfrac{1}{4}\) + x |= \(\dfrac{5}{6}\)

Help me..Thank you 💜

Answer 1

Answer 2

Ta có: \(\left|\dfrac{1}{4}+x\right|=\dfrac{5}{6}\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{12}-\dfrac{3}{12}\\x=\dfrac{-10}{12}-\dfrac{3}{12}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{12}\\x=\dfrac{-13}{12}\end{matrix}\right.\)

Vậy \(x=\dfrac{7}{12}\) hoặc \(x=-\dfrac{13}{12}\)

Answer 3

Ta có:

\(\Leftrightarrow\)\(\dfrac{1}{4}+x=\dfrac{5}{6}\) hoặc \(\dfrac{1}{4}+x=\dfrac{-5}{6}\)

\(\Rightarrow\)\(x=\dfrac{5}{6}-\dfrac{1}{4}\) \(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{7}{12}\) \(\Rightarrow x=\dfrac{-13}{12}\)

Vậy \(x=\dfrac{7}{12}\)hoặc \(x=\dfrac{-13}{12}\)