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\(\Rightarrow\left(2x-1\right)^3=\left(\frac{2}{3}\right)^3\)
\(\Rightarrow2x-1=\frac{2}{3}\)
biểu thức trở nên đơn giản r đó bn tự giải nha
tíc mình nha
(2x- 1)3 = \(\frac{8}{27}\)
(2x- 1)3= \(\frac{2^3}{3^3}\)
2x- 1= \(\frac{2}{3}\)
x= (\(\frac{2}{3}\)+1): 2
x= \(\frac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)
\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Leftrightarrow3x-1=3\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
b) Đề sai ! Sửa :
\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)
\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)
\(\Leftrightarrow x+3=\frac{4}{3}\)
\(\Leftrightarrow x=-\frac{5}{3}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)
\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)
\(\Leftrightarrow x=\frac{4}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x+\frac{1}{3}\right|+\frac{4}{5}=\left|-3,2+\frac{2}{5}\right|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{6}\right)...\left(27-\frac{3^5}{9}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}+\left(27-\frac{3^2}{6}\right)\left(27-\frac{3^3}{7}\right)...\left(27-27\right)...\left(27-\frac{3^{2010}}{2014}\right)\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|=2\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{3}=2\\x+\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-\frac{7}{3}\end{cases}}}\)
bạn ơi, có một chỗ chưa chuẩn .bạn kiểm tra lại giú mình. chỗ vế trái bạn thiếu \(\left(27-\frac{3}{5}\right)\). bạn bổ sung vào cho đúng nhé. dù sao vẫn cảm ơn bạn.
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e)
\(\left(x+3\right)^3=\left(x+3\right)^5\)
\(\Rightarrow\)\(x+3=1;0\)
TH1: TH2
\(x+3=0\) \(x+3=1\)
\(x=-3\) \(x=-2\)
\(x\in\left\{-3;-2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\left(2x+3\right)^2=\frac{3^2}{11^2}\)
từ đó suy ra
\(2x+3=\frac{3}{11}\)
2x=3/11-3
2x=-2/8/11
x=-2/8/11:2
x=-1/4/11
b,
(3x-1)^3=-8/27
(3x-1)^3=(-2/3)^3
Vậy suy ra
3x-1=-2/3
3x=-2/3+1
3x=1/3
x=1/3:3
x=1/9
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 \(x=-\frac{1}{2}\)