Question

tìm x biết :

1.\(\dfrac{-5}{6}+\dfrac{8}{3}+\dfrac{-29}{6}\le x\le\dfrac{-1}{2}+2+\dfrac{5}{2}\)

2.\(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{4}\)

3.\(\left|2x-\dfrac{3}{4}\right|=1\)

làm hộ mik 3 con này cái

Answer 1

2. \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{4}\)

\(\Leftrightarrow x.\left(\dfrac{27}{10}+\dfrac{-3}{2}\right)=\dfrac{-21}{4}.\dfrac{2}{7}\)

\(\Leftrightarrow x.\left(\dfrac{27}{10}+\dfrac{-15}{10}\right)=\dfrac{-3}{2}\)

\(\Leftrightarrow x.\dfrac{6}{5}=\dfrac{-3}{2}\)

\(\Leftrightarrow x=\dfrac{-3}{2}:\dfrac{6}{5}\)

\(\Leftrightarrow x=\dfrac{-3}{2}.\dfrac{5}{6}\)

\(\Leftrightarrow x=\dfrac{-5}{4}\)

Answer 2

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=\dfrac{-7}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}.\dfrac{1}{2}\\x=\dfrac{-7}{3}.\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{-7}{6}\end{matrix}\right.\)

vậy \(x\in\left\{\dfrac{7}{6};\dfrac{-7}{6}\right\}\)

Answer 3

1. bạn tính từ biểu thức ra rồi làm

dài dòng nên mình không tiện làm nhé