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Lời giải:
$0,25x-\frac{27}{8}x=\frac{-3}{4}$
$x(0,25-\frac{27}{8})=\frac{-3}{4}$
$x.\frac{-25}{8}=\frac{-3}{4}$
$x=\frac{-3}{4}: \frac{-25}{8}=\frac{6}{25}$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{2}{5}x-x=\frac{\left(-2018\right)^0}{5^2}\\ x\left(\frac{2}{5}-1\right)=\frac{1}{25}\\ x\left(\frac{2}{5}-\frac{5}{5}\right)=\frac{1}{25}\\ x\cdot\frac{-3}{5}=\frac{1}{25}\\ x=\frac{1}{25}:\frac{-3}{5}\\ x=\frac{1}{25}\cdot\frac{-5}{3}\\ x=\frac{-1}{15}\)Vậy \(x=\frac{-1}{15}\)
b) \(\left|-1\frac{1}{2}x+2x\right|-\frac{7}{4}=0,5\\ \left|x\left(-1\frac{1}{2}+2\right)\right|-\frac{7}{4}=\frac{1}{2}\\ \left|x\cdot\frac{1}{2}\right|=\frac{1}{2}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{2}{4}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x\cdot\frac{1}{2}=\frac{9}{4}\\x\cdot\frac{1}{2}=\frac{-9}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}:\frac{1}{2}\\x=\frac{-9}{4}:\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}\cdot2\\x=\frac{-9}{4}\cdot2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{2}\\x=\frac{-9}{2}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{9}{2};\frac{-9}{2}\right\}\)
c) \(x+\left(x+\frac{2}{7}\right)+\frac{-5}{11}=\frac{4}{11}\\ x+x+\frac{2}{7}=\frac{4}{11}-\frac{-5}{11}\\ 2x+\frac{2}{7}=\frac{4}{11}+\frac{5}{11}\\ 2x+\frac{2}{7}=\frac{9}{11}\\ 2x=\frac{9}{11}-\frac{2}{7}\\ 2x=\frac{63}{77}-\frac{22}{77}\\ 2x=\frac{41}{77}\\ x=\frac{41}{77}:2\\ x=\frac{41}{77\cdot2}\\ x=\frac{41}{154}\)Vậy \(x=\frac{41}{154}\)
d) \(\left|0,25x-20\%\right|+\frac{3}{8}=1\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\frac{3}{8}-\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\\ \Rightarrow\left[{}\begin{matrix}\frac{1}{4}x-\frac{1}{5}=1\\\frac{1}{4}x-\frac{1}{5}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=1+\frac{1}{5}\\\frac{1}{4}x=\left(-1\right)+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{5}{5}+\frac{1}{5}\\\frac{1}{4}x=\frac{-5}{5}+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{6}{5}\\\frac{1}{4}x=\frac{-4}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}:\frac{1}{4}\\x=\frac{-4}{5}:\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}\cdot4\\x=\frac{-4}{5}\cdot4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{24}{5}\\x=\frac{-16}{5}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{24}{5};\frac{-16}{5}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,0.25x\) - \(\frac{2}{3}x\) =\(1\frac{1}{6}\)
\(\Rightarrow x\cdot\left(0.25-\frac{2}{3}\right)=1\frac{1}{6}\)
\(\Rightarrow x\cdot\frac{-5}{12}=\frac{7}{6}\)
\(\Rightarrow x=\frac{7}{6}:\frac{-5}{12}\)
\(\Rightarrow x=\frac{-14}{5}=-2.8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(\left(x-2\right)\left(x+2\right)=0\)
\(< =>\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}< =>\orbr{\begin{cases}x=2\\x=-2\end{cases}}}\)
b,\(\left(\frac{3}{4}x-\frac{1}{2}\right)\left(0,25x+\frac{4}{3}\right)=0\)
\(< =>\orbr{\begin{cases}\frac{3x}{4}-\frac{2}{4}=0\\\frac{3x}{12}+\frac{16}{12}=0\end{cases}}\)
\(< =>\orbr{\begin{cases}3x-2=0\\3x+16=0\end{cases}}< =>\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{16}{3}\end{cases}}\)
\(\left(x-2\right)\left(x+2\right)=0\)
=> \(\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
\(\left(\frac{3}{4}x-\frac{1}{2}\right)\left(0,25x+\frac{4}{3}\right)=0\)
=> \(\orbr{\begin{cases}\frac{3}{4}x-\frac{1}{2}=0\\0,25x+\frac{4}{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{16}{3}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(0,25x+175\%x=x+\dfrac{9}{7}\\ x\left(0,25+1,75\right)=x+\dfrac{9}{7}\\ 2x=x+\dfrac{9}{7}\\ \dfrac{9}{7}=2x-x=x\\ x=\dfrac{9}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(0,25x+175\%x=x+\frac{9}{7}\)
\(0,25x+175\%x+x=\frac{9}{7}\)
\(0,25x+\frac{7}{4}x+x=\frac{9}{7}\)
\(\left(0,25+\frac{7}{4}+1\right)x=\frac{9}{7}\)
\(3x=\frac{9}{7}\)
\(x=\frac{9}{7}\div3\)
\(x=\frac{9}{7}\times\frac{1}{3}\)
\(x=\frac{3}{7}\)
vậy \(x=\frac{3}{7}\)
b)\(2x-\frac{3}{2}-x=1\frac{1}{5}\)
\(2x-x=1\frac{1}{5}+\frac{3}{2}\)
\(x=\frac{27}{10}\)
vậy \(x=\frac{27}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
0,25x -0,4x = 1
x(0,25 - 0,4) = 1
-0,15x =1
x = 1 : (-0,15)
x = - 20/3
![](https://rs.olm.vn/images/avt/0.png?1311)
0,25x12:0,15- (2,04+10,2:2,5):0,51
= 0,25x12:0,15- 6,12:0,51
= 3:0,15-12
= 20-12
= 8
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2\frac{1}{4}x-9\frac{1}{4}=20\)
\(\frac{9}{4}x=20+\frac{37}{4}\)
\(\frac{9}{4}x=\frac{80}{4}+\frac{37}{4}\)
\(\frac{9}{4}x=\frac{117}{4}\)
\(x=\frac{117}{4}:\frac{9}{4}\)
\(x=\frac{117}{4}.\frac{4}{9}\)
\(x=13\)
Vậy x=13
a) \(2\frac{1}{4}x-9\frac{1}{4}=20\)
\(\Rightarrow\frac{9}{4}x-\frac{37}{4}=20\)
\(\Rightarrow\frac{9}{4}x=20+\frac{37}{4}\)
\(\Rightarrow\frac{9}{4}x=\frac{117}{4}\)
\(\Rightarrow x=\frac{117}{4}:\frac{9}{4}\)
\(\Rightarrow x=13\)
Vậy x = 13
b) \(0,25x-\frac{1}{5}x=\frac{13}{20}\)
\(\Rightarrow\left(0,25-\frac{1}{5}\right)x=\frac{13}{20}\)
\(\Rightarrow\frac{1}{20}x=\frac{13}{20}\)
\(\Rightarrow x=\frac{13}{20}:\frac{1}{20}\)
\(\Rightarrow x=13\)
Vậy x = 13
x=0,5+0,25
x=0,75
x-0,25x=0,5
x(1-0,25)=0,5
x.0,75=0,5
x=0,5:0,75
x=\(\frac{2}{3}\)
Vậy x=\(\frac{2}{3}\)