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a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)
\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)
\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)
\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)
\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)
b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)
\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)
\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)
\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)
\(\Leftrightarrow x=\frac{1}{-3}\)
c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\1-2x=5\end{matrix}\right.\Leftrightarrow D\\ 2,\Leftrightarrow\left(-3\right)^x=-27\cdot81=-2187=\left(-3\right)^7\\ \Leftrightarrow x=7\left(A\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vũ Hồng Linh bạn check lại bài đầu dùm =_="
\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)
\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)
\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)
\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)
=> Làm nốt
Mấy bài kia cũng làm tương tự
(- \(\dfrac{1}{3}\))3.\(x\) = \(\dfrac{1}{81}\)
\(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))3
\(x\) = - (\(\dfrac{1}{3}\))4 :(\(\dfrac{1}{3}\))3
\(x=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x+5\right)^3=64\)
\(\Leftrightarrow\left(x+5\right)^3=4^3\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\)
Vậy x = - 1
b) \(x:\left(-\frac{3}{5}\right)^2=-\frac{3}{5}\)
\(\Leftrightarrow x=\left(-\frac{3}{5}\right)^2.\left(-\frac{3}{5}\right)\)
\(\Leftrightarrow x=\left(-\frac{3}{5}\right)^3\)
\(\Leftrightarrow x=-0,216\)
Vậy x = - 0, 216
c) \(\left(\frac{4}{7}\right)^4.x=\left(\frac{4}{7}\right)^6\)
\(\Leftrightarrow x=\left(\frac{4}{7}\right)^6:\left(\frac{4}{7}\right)^4\)
\(\Leftrightarrow x=\left(\frac{4}{7}\right)^2\)
\(\Leftrightarrow\text{x}=\frac{16}{49}\)
Vậy x = 16/49
d) \(\left(-\frac{1}{3}\right)^3x=\frac{1}{81}\)
\(\Leftrightarrow-\frac{1}{27}x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left(-\frac{1}{27}\right)\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy x = - 1/3
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(\left(x+2\right)^2=81>0\)
\(\orbr{\begin{cases}\left(x+2\right)^2=9^2\\\left(x+2\right)^2=\left(-9\right)^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+2=9\\x+2=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-11\end{cases}}\)
a)\(\left(x+2\right)^2=81\\ =>\left(x+2\right)=9^2hay\left(x+2\right)=\left(-9\right)^2\)
x+2=9 x+2=-9
x=9-2 x=-9-2
x=7(TM) x=-11(TM)
Vậy x=7 hay x=-9
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{\left(x-1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+7\right)}+....+\frac{1}{\left(x+76\right).\left(x+80\right)}=-\frac{81}{320}\)
\(\Rightarrow\frac{4}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x+3\right)\left(x+7\right)}+...+\frac{4}{\left(x+76\right)\left(x+80\right)}=\frac{-81}{80}\)
\(\Rightarrow\frac{1}{x-1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+7}+...+\frac{1}{x+76}-\frac{1}{x+80}=\frac{-81}{80}\)
\(\Rightarrow\frac{1}{x-1}-\frac{1}{x+90}=\frac{-81}{80}\)
Vì \(\frac{-81}{80}< 0\Rightarrow\frac{1}{x-1}< \frac{1}{x+90}\)
\(\Leftrightarrow x-1>x+90\)( luôn sai \(\forall x\in R\))
Vậy không tìm được x