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TT
29 tháng 10 2017
\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|\left(-3,2\right)+\dfrac{2}{5}\right|\)
\(\Leftrightarrow\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\dfrac{14}{5}\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|=\dfrac{14}{5}-\dfrac{4}{5}\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=2\\x-\dfrac{1}{3}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{-5}{3}\end{matrix}\right.\)
Vậy..............
GH
29 tháng 10 2017
\(\left | x - \frac{1}{3} \right | + \frac{4}{5} = \left | \left ( - 3,2 \right ) + \frac{2}{5}\right |\)
\(\left | x - \frac{1}{3} \right | + \frac{4}{5} = \left | - \frac{14}{5} \right |\)
\(\left | x - \frac{1}{3} \right | + \frac{4}{5} = \frac{14}{5} \)
\(\left | x - \frac{1}{3} \right | = 2\)
* \(x - \frac{1}{3}= 2\)
x = 2 + \( \frac{1}{3}\)
\(x = \frac{7}{3}\)
* \(x - \frac{1}{3}= - 2\)
\(x = - 2 + \frac{1}{3}\)
\(x = - \frac{5}{3}\)
Vậy x = \(x = \frac{7}{3}; x = - \frac{5}{3}\)
9 tháng 6 2017
\(\left\{\dfrac{-5< 0< -0,4}{x\in Z}\right\}\Rightarrow x\in\left\{-4;-3;-2;-1\right\}\)
11 tháng 2 2018
a) \(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\frac{-16}{5}+\frac{2}{5}\right|\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\frac{-14}{5}\right|\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)
\(\left|x-\frac{1}{3}\right|=\frac{14}{5}-\frac{4}{5}\)
\(\left|x-\frac{1}{3}\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{-5}{3}\end{cases}}\)
M
11 tháng 2 2018
làm tiếp câu a) nhé
b) \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{10}=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x-7=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=8\end{cases}}\)
1 tháng 12 2023
\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|-3,2+\dfrac{2}{5}\right|\)
=>\(\left|x-\dfrac{1}{3}\right|+0,8=\left|-3,2+0,4\right|=2,8\)
=>\(\left|x-\dfrac{1}{3}\right|=2,8-0,8=2\)
=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=2\\x-\dfrac{1}{3}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
2 tháng 9 2017
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 ...
BT
22 tháng 2 2018
Ta có: \(\left|x-\dfrac{1}{3}\right|\)+0,8 = \(\left|-3,2+0,4\right|\)
---> \(\left|x-\dfrac{1}{3}\right|\)+0,8 =2,8 ---> \(\left|x-\dfrac{1}{3}\right|\)= 2,8-0,8=2 --> TH1: x-\(\dfrac{1}{3}\) = 2 --> x= 2+ \(\dfrac{1}{3}\)= \(\dfrac{7}{3}\) TH2: x-\(\dfrac{1}{3}\) = -2---> x= -2+\(\dfrac{1}{3}\)=\(\dfrac{-5}{3}\) Vạy x =\(\dfrac{-5}{3}\) hoặc x= \(\dfrac{7}{3}\) Tick cho mik nha!!
11 tháng 2 2018
a) \(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|\left(-3,2\right)+\dfrac{2}{5}\right|\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|+0,8=\left|-3,2+0,4\right|\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|+0,8=2,8\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=2\\x-\dfrac{1}{3}=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{-5}{3}\end{matrix}\right.\)