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\(\dfrac{2x-1}{3}=\dfrac{7}{27\left(2x+1\right)}\)
\(\Leftrightarrow27\left(4x^2-1\right)=21\)
\(\Leftrightarrow4x^2-1=\dfrac{7}{9}\)
\(\Leftrightarrow4x^2=\dfrac{16}{9}\)
\(\Leftrightarrow x^2=\dfrac{4}{9}\)
=>x=2/3 hoặc x=-2/3
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)
\(\frac{1}{2}-x=\frac{57}{28}\)
\(x=-\frac{43}{28}\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow\left(2x-1\right)^2=5^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy ...
a) \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=\frac{-11}{4}\)
\(\Rightarrow\left(\frac{1}{2}-x\right)=\left(-\frac{5}{7}\right)+\frac{11}{4}\)
\(\Rightarrow\frac{1}{2}-x=\frac{57}{28}\)
\(\Rightarrow x=\frac{1}{2}-\frac{57}{28}\)
\(\Rightarrow x=-\frac{43}{28}\)
Vậy \(x=-\frac{43}{28}.\)
b) \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=20+5\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=5+1=6\\2x=\left(-5\right)+1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6:2\\x=\left(-4\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{3;-2\right\}.\)
d) \(\frac{x-6}{4}=\frac{4}{x-6}\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=4.4\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=16\)
\(\Rightarrow\left(x-6\right)^2=16\)
\(\Rightarrow x-6=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}x-6=4\\x-6=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+6\\x=\left(-4\right)+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{10;2\right\}.\)
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
(-1/3)^2x-1=-1/27
=>-1^2x-1/3^2x-1=1/27
Mà -1^2x-1=(-1)^2x/(-1)^1
=1/-1=-1
=>3^2x-1=27
=>3^2x/3^1=27
=>3^2x/3=27
=>3^2x=27*3
=>3^2x=81
=>3^2x=3^4
=>2x=4
=>x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-1}{4}\) \(\Leftrightarrow\frac{2x-2}{10}=\frac{3y-6}{9}=\frac{2z-2}{8}\)
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{2x-2}{10}=\frac{3y-6}{9}=\frac{2z-2}{8}=\frac{2x-2-3y+6-2z+2}{10-9-8}=\frac{-27+6}{-7}=\frac{-21}{-7}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x-1}{5}=3\\\frac{y-2}{3}=3\\\frac{z-1}{4}=3\end{cases}\Rightarrow}\hept{\begin{cases}x-1=15\\y-2=9\\z-1=12\end{cases}\Rightarrow}\hept{\begin{cases}x=16\\y=11\\z=13\end{cases}}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(2x+\frac{1}{3}\right)^3=\frac{-8}{27}\)
\(\Leftrightarrow2x+\frac{1}{3}=\frac{-2}{3}\)
\(\Leftrightarrow2x=\frac{-2}{3}-\frac{1}{3}=-1\)
hay \(x=-\frac{1}{2}\)
Vậy: \(x=-\frac{1}{2}\)
\(\left(2x+\frac{1}{3}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(2x+\frac{1}{3}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow2x+\frac{1}{3}=-\frac{2}{3}\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\)
\(\frac{2x-1}{3}=\frac{-27}{1-2x}\)
\(\Rightarrow\left(2x-1\right)\left(1-2x\right)=\left(-27\right).3\)
\(\Rightarrow2x-4x^2-1-2x=-81\)
\(\Rightarrow4x^2-1=-81\)
\(\Rightarrow4x^2=-81+1\)
\(\Rightarrow4x^2=-80\)
Vì \(x^2\ge0\)mà \(4x^2=-80\)
\(\Rightarrow x\in\theta\)
tíc mình nha