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\(\frac{x}{-15}=-\frac{60}{x}\)
\(\Rightarrow x^2=-15.\left(-60\right)\)
\(\Rightarrow x^2=900\)
\(\Rightarrow x=\sqrt{900}\)
\(x=30\)
\(\frac{x}{-15}=\frac{-60}{x}\)
\(\Rightarrow x\cdot x=\left(-60\right)\cdot\left(-15\right)\)
\(\Rightarrow x^2=900\)
=>x2=(-30)2 hoặc 302
=>x=30 hoặc -30
![](https://rs.olm.vn/images/avt/0.png?1311)
1\(x\) - (\(x\) - \(\dfrac{1}{3}\)) = \(\dfrac{1}{6}\)
\(x\) - \(x\) + \(\dfrac{1}{3}\) = \(\dfrac{1}{6}\)
\(\dfrac{1}{3}\) = \(\dfrac{1}{6}\) (vô lí)
Vậy không có giá trị nào của \(x\) thỏa mãn đề bài
\(\dfrac{x-1}{-15}\) = - \(\dfrac{60}{x-1}\)
(\(x\) - 1).(\(x\) - 1) = (-60).(-15)
(\(x\) - 1)2 = 900
\(\left[{}\begin{matrix}x-1=-30\\x-1=30\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-29\\x=31\end{matrix}\right.\)
Vậy \(x\in\) {-29; 31}
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\(\frac{X}{-15}\)=\(\frac{-60}{x}\)
=> \(^{x^2}\)=-15.(-60)
\(x^2\)=900
\(x^2\)=30.30
\(x^2\)=\(30^2\)
=> x=30
![](https://rs.olm.vn/images/avt/0.png?1311)
x-1/-15 = -60/x-1
=> (x-1)2 = 900
=> (x-1)2 = 302
=> \(\hept{\begin{cases}x-1=-30\\x-1=30\end{cases}}\)
=>\(\hept{\begin{cases}x=-29\\x=31\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(-\frac{x}{15}=-\frac{60}{x}\Rightarrow-15.\left(-60\right)=x^2\Rightarrow900=x^2\Rightarrow x=\sqrt{900}=30\)
\(x=-\sqrt{900}=-30\)
b) \(-\frac{2}{x}=-\frac{x}{0,32}\Rightarrow-2.\left(-0,32\right)=x^2\Rightarrow\frac{16}{25}=x^2\Rightarrow x=\frac{\sqrt{16}}{\sqrt{25}}=\frac{4}{5}\)
\(x=-\frac{\sqrt{16}}{\sqrt{25}}=-\frac{4}{5}\)
a) \(\frac{x}{-15}=\frac{-60}{x}\)
=> x.x = (-60) . (-15)
=>x2 = 900
=>x2 = 302
=>x=30
b)\(\frac{-2}{x}=\frac{-x}{0,32}\)
=>x.(-x) =(-2).0,32
=>-(x2) =-0,64
=>-(x2) =-(0,82)
=>x=0,8
![](https://rs.olm.vn/images/avt/0.png?1311)
a)x/-15=-60/x
<=> x.x=-15.(-60)
2x=900
x=900:2
x=450
vậy x=450
b) -2/x=-x/8
x.(-x)=-2.8
-2x=-16
x=-16:(-2)
x=8
vậy x=8
a) Ta có: x /-15 = -60/x <=> (-15)(-60)= x.x
<=> 900= x^2 => x= -30; 30
b) Ta có : (-2)*8 = x*(-x)
<=> -16= x*(-x) <=> 16= x^2
=> x= -4; 4
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{x}{-15}=-\frac{60}{x}\)
\(\Rightarrow x.x=\left(-60\right).\left(-15\right)\)
\(\Rightarrow x^2=900=30^2\)
\(\Rightarrow x=30\)
b) \(-\frac{2}{x}=-\frac{x}{\frac{8}{25}}\)
\(\Rightarrow x.x=-2.\frac{-8}{25}\)
\(\Rightarrow x^2=\frac{16}{25}=\left(\frac{4}{5}\right)^2\)
\(\Rightarrow x=\frac{4}{5}\)
\(a,\frac{x}{-15}=\frac{-60}{x}\)
\(\Rightarrow x.x=\left(-60\right).\left(-15\right)\)
\(\Rightarrow x^2=900\)
\(\Rightarrow x^2=\orbr{\begin{cases}30^2\\-30^2\end{cases}}\)
\(\Rightarrow x=\orbr{\begin{cases}30\\-30\end{cases}}\)
\(b,\frac{-2}{x}=\frac{-x}{8}:25\)
\(\Rightarrow-x^2=-2.8.25=-400\)
\(\Rightarrow x^2=400\)
\(\Rightarrow x=\orbr{\begin{cases}20\\-20\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x}{-15}=\frac{-60}{x}\Rightarrow x.x=\left(-60\right).\left(-15\right)\)
\(x^2=900\)
\(\Rightarrow x=30\)hoặc \(x=-30\)
Ta có: \(\frac{x}{-15}=\frac{-60}{x}\)
=>-15.(-60)=x.x
=>900=x2
=>x2=302=(-30)2
=>x=30 hoặc x=-30
Vậy x=30 hoặc x=-30