Tìm x biết: |x - 3,2| + |2x - \(\frac{1}{5}\)| = x + 3.
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\(\left|x-3,2\right|+\left|2x-\frac{1}{5}\right|=x+3.\)
ĐK : \(x+3\ge0\Leftrightarrow x\ge-3\)
Th1 : \(x-3,2+2x-\frac{1}{5}=x+3\)
\(x-3,2+2x=x+\frac{16}{5}\)
\(x+2x=x+\frac{32}{5}\)
\(2x=\frac{32}{5}\)
\(\Leftrightarrow x=3,2\)(tm)
\(x-3,2+2x-\frac{1}{5}=3-x\)
\(x-3,2+2x=3-x+\frac{1}{5}\)
\(x-3,2+2x=\frac{16}{5}-x\)
\(x+2x=\frac{16}{5}-x+3,2\)
\(x+2x=\frac{32}{5}-x\)
\(2x=\frac{32}{5}-x-x\)
\(2x=\frac{32}{5}-2x\)
\(4x=\frac{32}{5}\)
\(x=1,6\)(tm)
Vậy \(x=1,6\)hoặc \(x=3,2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Bài 3:
a: \(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\\dfrac{3}{4}x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}3x+2>0\\\dfrac{2}{3}x-5< 0\end{matrix}\right.\Leftrightarrow-\dfrac{2}{3}< x< \dfrac{15}{2}\)
c: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x+2=0\\\dfrac{2}{5}x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{3}{4}=-2\\\dfrac{2}{5}x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{8}{3}\\x=6:\dfrac{2}{5}=15\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)
\(\Rightarrow\left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)
\(\Rightarrow\left|x-\frac{1}{3}\right|=2\)
\(\Rightarrow x-\frac{1}{3}=\hept{\begin{cases}2\\-2\end{cases}}\)
\(\Rightarrow x=\hept{\begin{cases}\frac{7}{3}\\\frac{-5}{3}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|-2,8\right|\)
\(\left|x-\frac{1}{3}\right|=2,8-\frac{4}{5}\)
\(\left|x-\frac{1}{3}\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{-5}{3}\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{5}{3}\end{cases}}\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)
\(\left|x-\frac{1}{3}\right|=\frac{14}{5}+\frac{4}{5}\)
\(\left|x-\frac{1}{3}\right|=\frac{19}{5}\)
\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{3}=\frac{19}{5}\Leftrightarrow x=\frac{62}{15}\\x-\frac{1}{3}=-\frac{19}{5}\Leftrightarrow x=-\frac{52}{15}\end{cases}}\)
vây x= \(\frac{62}{15}\)hoặc x=\(-\frac{52}{15}\)
Chúc bn hok tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,|x-\frac{1}{3}|+\frac{4}{5}=|-3,2+\frac{2}{5}|\)
\(\Rightarrow|x-\frac{1}{3}|+\frac{4}{5}=|-\frac{7}{10}|\)
\(\Rightarrow|x-\frac{1}{3}|+\frac{4}{5}=\frac{7}{10}\)
\(\Rightarrow|x-\frac{1}{3}|=\frac{7}{10}-\frac{4}{5}\)
\(\Rightarrow|x-\frac{1}{3}|=-\frac{1}{10}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{10}\\x-\frac{1}{3}=-\frac{1}{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{13}{30}\\x=\frac{7}{30}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
/2x-1/=12
=) 2x-1=12
2x=13
x=13/2
hoặc
2x-1=(-15)+(-3)
2x-1=-18
2x=-17
x=-17/2
duyệt nha
![](https://rs.olm.vn/images/avt/0.png?1311)
|x-1/3|+4/5=14/5
|x-1/3|=2
=>x-1/3=2 hoặc x-1/3=-2
=>x=7/3 hoặc x=-5/3
vậy x=7/3 hoặc x=-5/3
tk mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (2x-1)\(^2\)+\(\left|2y-x\right|\)=0
Vì (2x-1)\(^2\)\(\ge\)0 với mọi x
\(\left|2y-x\right|\)\(\ge\)0 với mọi y
\(\Rightarrow\)\(\left\{\begin{matrix}2x-1=0\\2y-x=0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{\begin{matrix}x=\frac{1}{2}\\2y-\frac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)
Vậy .....
b)\(\left|x-\frac{1}{3}\right|\)+\(\frac{4}{5}\)=\(\frac{14}{5}\)
\(\Rightarrow\)\(\left|x-\frac{1}{3}\right|\)=2
\(\Rightarrow\)\(\left[\begin{matrix}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{matrix}\right.\)\(\Rightarrow\)\(\left[\begin{matrix}x=\frac{7}{3}\\x=\frac{-5}{3}\end{matrix}\right.\)
Vậy ....
đây nhé