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Tìm x :
\(55^{2x}:11^{2x}=625\)
\(\Leftrightarrow\left(55:11\right)^{2x}=625\)
\(\Leftrightarrow5^{2x}=625\)
\(\Leftrightarrow5^{2x}=5^4\)
\(\Rightarrow2x=4\)
\(\Leftrightarrow x=4:2\)
\(\Leftrightarrow x=2\)
Vậy x =2
\(55^{2x}:11^{2x}=625\)
\(55^{2x}\cdot\left(\frac{1}{11}\right)^{2x}=625\)
\(\left(55\cdot\frac{1}{11}\right)^{2x}=625\)
\(5^{2x}=5^4\)
\(\Leftrightarrow2x=4\)
\(x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)
\(\Leftrightarrow\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-\sqrt{\dfrac{9}{4}}\right)^2=\dfrac{1}{\sqrt{625}}\)
\(\Leftrightarrow\left(2x-\dfrac{3}{2}\right)^2=\dfrac{1}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{2}=\sqrt{\dfrac{1}{25}}\\2x-\dfrac{3}{2}=-\sqrt{\dfrac{1}{25}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{2}=\dfrac{1}{5}\\2x-\dfrac{3}{2}=-\dfrac{1}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{10}\\2x=\dfrac{13}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{20}\\x=\dfrac{13}{20}\end{matrix}\right.\)
Vậy ..
\(\left(2x-\sqrt{\dfrac{9}{4}}\right)^2=\dfrac{1}{\sqrt{625}}\\ \Leftrightarrow\left(2x-\dfrac{3}{2}\right)^2=\dfrac{1}{25}\\ \Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{2}=\dfrac{1}{5}\\2x-\dfrac{3}{2}=-\dfrac{1}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{20}\\x=\dfrac{13}{20}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-\sqrt{\dfrac{9}{4}}\right)^2=\dfrac{1}{\sqrt{625}}\)
\(\Leftrightarrow\left(2x-\dfrac{3}{2}\right)^2=\dfrac{1}{25}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{2}=\dfrac{1}{5}\\2x-\dfrac{3}{2}=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{20}\\x=\dfrac{13}{20}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\\ \Rightarrow\left(x-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow x-\dfrac{2}{15}=\dfrac{2}{5}\\ \Rightarrow x=\dfrac{2}{5}+\dfrac{2}{15}\\ \Rightarrow x=\dfrac{6}{15}+\dfrac{2}{15}\\ \Rightarrow x=\dfrac{8}{15}\\ \left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\\ \Rightarrow\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\\ \Rightarrow2x+5=4\\ \Rightarrow2x=4-5\\ \Rightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)
\(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\)
\(\left(x-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\)
\(x-\dfrac{2}{15}=\dfrac{2}{5}\)
\(x=\dfrac{2}{5}+\dfrac{2}{15}\)
\(x=\dfrac{8}{15}\)
\(\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)
\(\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\)
\(2x+5=4\)
\(2x=-1\)
\(x=-0,5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(55^{2x}\div11^{2x}=625\)
\(\left(55:11\right)^{2x}=625\)
\(5^{2x}=5^4\Rightarrow2x=4\Rightarrow x=2\)
Ta có \(55^{2x}\div11^{2x}=625\)
\(\Rightarrow\left(55\div11\right)^{2x}=625\)
\(\Rightarrow5^{2x}=5^4\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)