`(3/5-x) . 7/11 = -21/22`
`3/5 - x = -21/22 : 7/11`
`3/5 - x = -21/22 xx 11/7`
`3/5 - x = -3/2`
`x=3/5-(-3/2)`
`x=3/5+3/2`
`x=6/10 + 15/10`
`x=21/10`
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`x/25 = 1/x`
`x.x=25.1`
`x^2 = 25`
`=> x^2`\(=\left(\pm5\right)\)
`=> x = {5;-5}`
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`|x - 1/9| = 4/9`
`@TH1:`
`x-1/9=4/9`
`x=4/9+1/9`
`x=5/9`
`@TH2:`
`x-1/9=-4/9`
`x=-4/9+1/9`
`x=-3/9`
`x=-1/3`
Vậy `x = {-1/3; 5/9}`
\(a)\left(\dfrac{3}{5}-x\right).\dfrac{7}{11}=-\dfrac{21}{22}\\ \left(\dfrac{3}{5}-x\right)=-\dfrac{21}{22}:\dfrac{7}{11}\\ \left(\dfrac{3}{5}-x\right)=-\dfrac{3}{2}\\ x=\dfrac{3}{5}--\dfrac{3}{2}\\ x=\dfrac{21}{10}\\ b)\dfrac{x}{25}=\dfrac{1}{x}\\ x^2=25\\ x=\sqrt{25}\\ x=5\\c)\left|x-\dfrac{1}{9}\right|=\dfrac{4}{9}\\ x-\dfrac{1}{9}=\dfrac{4}{9}\\ x=\dfrac{4}{9}+\dfrac{1}{9}\\ x=\dfrac{5}{9}. \)
a) \(\left(\dfrac{3}{5}-x\right).\dfrac{7}{11}=-\dfrac{21}{22}\) \(\Rightarrow\dfrac{3}{5}-x=-\dfrac{3}{2}\Rightarrow x=\dfrac{21}{10}\)
b) \(\dfrac{x}{25}=\dfrac{1}{x}\left(x\ne0\right)\Rightarrow x^2=25\Rightarrow x=\pm5\)
c) \(...\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{9}=\dfrac{4}{9}\\x-\dfrac{1}{9}=-\dfrac{4}{9}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{9}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
a) \(\left(\dfrac{3}{5}-x\right)\cdot\dfrac{7}{11}=-\dfrac{21}{22}\)
\(\left(\dfrac{3}{5}-x\right)=\left(-\dfrac{21}{22}\right):\dfrac{7}{11}\)
\(\dfrac{3}{5}-x=-\dfrac{3}{2}\)
\(x=\dfrac{3}{5}+\dfrac{3}{2}\)
`=>x=21/10`
b) \(\dfrac{x}{25}=\dfrac{1}{x}\)
\(x\cdot x=1\cdot25\)
\(x^2=25\)
\(x^2=5^2\)
`=>`\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
c) \(\left|x-\dfrac{1}{9}\right|=\dfrac{4}{9}\)
\(x-\dfrac{1}{9}=\dfrac{4}{9}\)
\(\left[{}\begin{matrix}x-\dfrac{1}{9}=\dfrac{4}{9}\\x-\dfrac{1}{9}=-\dfrac{4}{9}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{5}{9}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
