x-0,25x=0,5
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![](https://rs.olm.vn/images/avt/0.png?1311)
x 3 – 0,25x = 0
⇔x( x 2 - 0,25) = 0
⇔ x( x 2 - 0 , 5 2 ) = 0
⇔ x(x + 0,5)(x – 0,5) = 0
+) x + 0,5 = 0 ⇔ x= - 0,5
+) x - 0,5 = 0 ⇔ x = 0,5
Vậy x= 0, x= - 0,5; x= 0,5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-0,25=x^2-\frac{1}{4}=x\left(x-\frac{1}{4}\right)\)
~ Hok tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow x\left(x^2-0,25\right)=0\\ \Rightarrow x\left(x-0,5\right)\left(x+0,5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-0,5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
Bg
x3 - 0,25x = 0
x2.x - 0,25x = 0
x(x2 - 0,25) = 0
=> x = 0 hoặc x2 - 0,25 = 0
Với x2 - 0,25 = 0:
x2 = 0,25
\(\sqrt{x^2}=\sqrt{0,25}\)
x = \(\frac{1}{2}\)
Vậy x = 0 và x = \(\frac{1}{2}\)
x3 - 0, 25x = 0
<=> x( x2 - 0, 25 ) = 0
<=> x( x2 - 1/4 ) = 0
<=> x( x - 1/2 )( x + 1/2 ) = 0
<=> x = 0 hoặc x - 1/2 = 0 hoặc x + 1/2 = 0
<=> x = 0 hoặc x = ±1/2
![](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)
x^3-0,25x^2=0
<=>x2.(x-0,25)=0
<=>x=0 hoặc x-0,25=0
<=>x=0 hoặc x=0,25
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-0,25x=0\)
\(x\left(x^2-0,25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-0,25=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=0,25\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{1}{2}\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=\pm\frac{1}{2}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-0,25x=0\)
⇔\(x\left(x-0,25\right)=0\)
⇔ x = 0 hoặc x = 0,25
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=\(\frac{2}{3}\)