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P/s : mk làm từng phần một
\(\left(\frac{2}{3}\right)^x=\frac{8}{27}=\left(\frac{2}{3}\right)^3\)
=> x = 3
Vậy,........
2)
\(\left(\frac{5}{6}x+\frac{1}{2}\right)^2=\frac{9}{16}=\left(\frac{3}{4}\right)^2\)
\(\frac{5}{6}x+\frac{1}{2}=\frac{3}{4}\)
\(\frac{5}{6}x=\frac{1}{4}\)
\(x=\frac{3}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.a) \(\left(\frac{3}{5}\right)^{15}:\left(\frac{9}{25}\right)^5=\frac{3^{15}}{5^{15}}.\frac{5^{10}}{3^{10}}=\frac{3^5}{5^5}=\left(\frac{3}{5}\right)^5\)
b)\(\left(\frac{2}{3}\right)^{10}:\left(\frac{4}{9}\right)^4=\frac{2^{10}}{3^{10}}.\frac{3^8}{2^8}=\frac{2^2}{3^2}=\left(\frac{2}{3}\right)^2\)
2.
a)\(2^x=4\Rightarrow2^x=2^2\Rightarrow x=2\)
b)\(x^3=-27\Rightarrow x^3=-3^3\Rightarrow x=-3\)
c)\(x^2=16\Rightarrow x=\pm4\)
d)\(\left(x+1\right)^2=9\Rightarrow\hept{\begin{cases}x+1=3\Rightarrow x=2\\x+1=-3\Rightarrow x=-4\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 5^2x-1 = 125
=> 5^2x-1 = 5^3
=> 2x - 1 = 3
=> 2x = 3 + 1
=> 2x = 4
=> x = 4 : 2 = 2
b) 3^x+3 - 3^x+2 - 3^x = 153
3^x . 3^3 - 3^x . 3^2 - 3^x = 153
=> 3^x . (3^3 - 3^2 - 1) = 153
=> 3^x . (27 - 9 - 1) = 153
=> 3^x . 17 = 153
=> 3^x = 153 : 17
=> 3^x = 9
3^x = 3^2
=> x = 2
Sửa đề: \(\dfrac{16}{15}\rightarrow\dfrac{16}{25}\)
Giải:
\(\left(x-\dfrac{3}{5}\right)^2=\dfrac{16}{25}\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-\dfrac{3}{5}\right)^2=\left(\dfrac{4}{5}\right)^2\\\left(x-\dfrac{3}{5}\right)^2=\left(\dfrac{-4}{5}\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{5}=\dfrac{4}{5}\\x-\dfrac{3}{5}=\dfrac{-4}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{-1}{5}\end{matrix}\right.\)
\(\left(\dfrac{5}{3}-x\right)^3=\dfrac{1}{27}\)
\(\Rightarrow\left(\dfrac{5}{3}-x\right)^3=\left(\dfrac{1}{3}\right)^3\)
\(\dfrac{5}{3}-x=\dfrac{1}{3}\)
\(x=\dfrac{5}{3}-\dfrac{1}{3}\)
\(x=\dfrac{4}{3}\)