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a) Có: 2\(^5\) = 32
Nên 2\(^{x+1}\)= 32
Nên x+1 = 5
x = 5-1
x = 4
b) Hình như sai đề, bạn xem lại thử nha
c) (7\(^x\))\(^2\)= 7\(^{14}\)
7\(^x\) = 7\(^7\)
x = 7
d) Cái này cũng hơi có vấn đề này. Vì (-0.5)\(^5\)= \(\frac{-1}{32}\) mà xem lại nha!!

a,\(\left(x-\frac{7}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)
\(x-\frac{7}{9}=\frac{4}{9}\)
\(x=\frac{4}{9}+\frac{7}{9}\)
\(x=\frac{11}{9}\)
Vậy x=\(\frac{11}{9}\)

a) \(\left(\frac{16}{2}\right)^x=2\)
\(8^x=2\)
<=> 23x = 21
<=> 3x = 1
=> x = \(\frac{1}{3}\)
b) 8x : 2x = 4
(8 : 2)x = 4
4x = 4
=> x = 1
c) \(\left(\frac{1}{2}\right)^x=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\)
=> x = 5

\(\frac{64}{\left(-2\right)^x}=-32\)
\(\Rightarrow64\div\left(-2\right)^x=-32\)
\(\Rightarrow\left(-2\right)^x=64\div\left(-32\right)\)
\(\Rightarrow\left(-2\right)^x=\left(-2\right)^1\)
\(\Rightarrow x=1\)
\(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)
Xét \(\left(x-2\right)\left(x+\frac{2}{3}\right)=0\Rightarrow\hept{\begin{cases}x-2=0\\x+\frac{2}{3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\x=-\frac{2}{3}\end{cases}}\)
Mà \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\) nên \(\left(x-2\right)\) và \(\left(x+\frac{2}{3}\right)\) đồng dấu
Suy ra \(\hept{\begin{cases}x>2\\x>-\frac{2}{3}\end{cases}\Leftrightarrow x>2}\) (do \(2>-\frac{2}{3}\)) hoặc \(\hept{\begin{cases}x< 2\\x< -\frac{2}{3}\end{cases}\Leftrightarrow x< -\frac{2}{3}}\) (do \(-\frac{2}{3}< 2\))
Vậy \(\hept{\begin{cases}x>2\\x< -\frac{2}{3}\end{cases}}\)
Lưu ý rằng: dấu ngoặc \(\hept{\begin{cases}...\\...\end{cases}}\) thay thế cho chữ hoặc nhé!

a. 2x-1+ 5.2x-1:2=7/32
=> 2x+1.(1+5/2)=7/32
=>2x+1.7/2=7/32
=> 2x+1=1/16=1/24
=> x+1=-4=>x=-5

1) \(\left(\frac{2x}{3}-3\right):\left(-10\right)=\frac{2}{5}\)
\(\Leftrightarrow-\frac{\frac{2x}{3}-3}{10}=\frac{2}{5}\)
\(\Leftrightarrow-\left(\frac{\frac{2x}{3}}{10}-\frac{3}{10}\right)=\frac{2}{5}\)
\(\Leftrightarrow-\left(\frac{2x}{3\times10}-\frac{3}{10}\right)=\frac{2}{5}\)
\(\Leftrightarrow-\left(\frac{2x}{30}-\frac{3}{10}\right)=\frac{2}{5}\)
\(\Leftrightarrow-\frac{x}{15}+\frac{3}{10}=\frac{2}{5}\)
\(\Leftrightarrow\frac{3}{10}-\frac{x}{15}=\frac{2}{5}\)
\(\Leftrightarrow-\frac{x}{15}=\frac{2}{5}-\frac{3}{10}\)
\(\Leftrightarrow-\frac{x}{15}=\frac{1}{10}\)
\(\Leftrightarrow-x=\frac{15}{10}\)
\(\Leftrightarrow-x=\frac{3}{2}\)
\(\Leftrightarrow x=-\frac{3}{2}\)
Vậy \(x=-\frac{3}{2}\)
2) \(\left|2x-1\right|+1=4\)
\(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
Vậy \(x\in\left\{2;-1\right\}\)

a.
\(\left(\frac{1}{3}\right)^2\times27=3^x\)
\(\frac{1^2}{3^2}\times3^3=3^x\)
\(3^1=3^x\)
\(x=1\)
b.
\(\frac{64}{\left(-2\right)^x}=-32\)
\(\frac{\left(-2\right)^6}{\left(-2\right)^x}=\left(-2\right)^5\)
\(\left(-2\right)^x=\frac{\left(-2\right)^6}{\left(-2\right)^5}\)
\(\left(-2\right)^x=-2\)
\(x=1\)
c.
\(3x^2-\frac{1}{2}x=0\)
\(x\times\left(3x-\frac{1}{2}\right)=0\)
TH1:
\(x=0\)
TH2:
\(3x-\frac{1}{2}=0\)
\(3x=\frac{1}{2}\)
\(x=\frac{1}{2}\div3\)
\(x=\frac{1}{2}\times\frac{1}{3}\)
\(x=\frac{1}{6}\)
Vậy x = 0 hoặc x = 1/6
\(2\left(x+1\right)^2=32\)
\(\Leftrightarrow\left(x+1\right)^2=32:2=16\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=4\\x+1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}}\)
(x+1)2 = 32/2 =16 = 42
th1: x+1 = 4
x = 3
th2; x+1 = -4
x = -5