dề bài đâu mà tìm?

a) \(5^x-\left(5^3\right)^2=625\)

\(\Leftrightarrow5^x-5^6=5^4\)

\(\Leftrightarrow x-6=4\)

\(\Leftrightarrow x=10\)

b) \(\left(\frac{12}{25}\right)^x=\left(\frac{5}{3}\right)^{-2}-\left(-\frac{3}{5}\right)^4\)

\(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\)

\(\left(x-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\)

\(x-\dfrac{2}{15}=\dfrac{2}{5}\)

\(x=\dfrac{2}{5}+\dfrac{2}{15}\)

\(x=\dfrac{8}{15}\)

\(\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)

\(\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\)

\(2x+5=4\)

\(2x=-1\)

\(x=-0,5\)

\(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\\ \Rightarrow\left(x-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow x-\dfrac{2}{15}=\dfrac{2}{5}\\ \Rightarrow x=\dfrac{2}{5}+\dfrac{2}{15}\\ \Rightarrow x=\dfrac{6}{15}+\dfrac{2}{15}\\ \Rightarrow x=\dfrac{8}{15}\\ \left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\\ \Rightarrow\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\\ \Rightarrow2x+5=4\\ \Rightarrow2x=4-5\\ \Rightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)

5^x.5^6=5^3

5^x.(5³)²=625

=> 5^x.5⁶=5⁴

=> 5^x=5⁴:5⁶

=> 5^x=5^-2

=> x=-2

Vậy x=-2

\(^{=>5^x.5^6=5^4}\)

=> \(5^x=5^{4-6}\)

=>x=-2

\(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\left(\dfrac{5}{3}\right)^6:\left(\dfrac{5^4}{7^3}\right)^2\)

\(=\dfrac{3^6}{7^6}\cdot\dfrac{5^6}{3^6}:\dfrac{5^8}{7^6}\)

\(=\dfrac{5^6}{5^8}=\dfrac{1}{25}\)

d,\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\\ \Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\\ \Leftrightarrow x-\frac{2}{9}=\frac{4}{9}\\ \Leftrightarrow x=\frac{6}{9}\)

Vậy...

a) \(\left(x-3\right).\left(4-5x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\4-5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+3\\5x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=4:5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\frac{4}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{3;\frac{4}{5}\right\}.\)

b) \(\left|x+\frac{3}{4}\right|+\frac{1}{3}=0\)

\(\Rightarrow\left|x+\frac{3}{4}\right|=0-\frac{1}{3}\)

\(\Rightarrow\left|x+\frac{3}{4}\right|=-\frac{1}{3}.\)

Ta luôn có: \(\left|x\right|\ge0\) \(\forall x.\)

\(\Rightarrow\left|x+\frac{3}{4}\right|>-\frac{1}{3}\)

\(\Rightarrow\left|x+\frac{3}{4}\right|\ne-\frac{1}{3}.\)

Vậy \(x\in\varnothing.\)

c) \(5^x.\left(5^3\right)^2=625\)

\(\Rightarrow5^x.5^6=5^4\)

\(\Rightarrow5^{x+6}=5^4\)

\(\Rightarrow x+6=4\)

\(\Rightarrow x=4-6\)

\(\Rightarrow x=-2\)

Vậy \(x=-2.\)

Chúc bạn học tốt!

a) (x - 3) (4 - 5x) = 0

\(\Rightarrow\) \(\left[{}\begin{matrix}x-3=0\\4-5x=0\end{matrix}\right.\)

\(\Rightarrow\) \(\left[{}\begin{matrix}x=3\\5x=4\end{matrix}\right.\)

\(\Rightarrow\) \(\left[{}\begin{matrix}x=3\\x=0,8\end{matrix}\right.\)

Vậy x \(\in\) \(\left\{3;0,8\right\}\)

b) |x + \(\frac{3}{4}\) | + \(\frac{1}{3}\) = 0

|x + \(\frac{3}{4}\) | = \(\frac{1}{3}\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{3}\\x+\frac{3}{4}=\frac{-1}{3}\end{matrix}\right.\)

\(\Rightarrow\) \(\left[{}\begin{matrix}x+\frac{9}{12}=\frac{4}{12}\\x+\frac{9}{12}=\frac{-4}{12}\end{matrix}\right.\)

\(\Rightarrow\) \(\left[{}\begin{matrix}x=\frac{-5}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)

Vậy x \(\in\) \(\left\{\frac{-5}{12};\frac{-13}{12}\right\}\)

c) 5\(^x\) . (5\(^3\))\(^2\) = 625

5\(^x\) . 5\(^6\) = 5\(^4\)

x + 6 = 4

x = 4 - 6

x = -2

Vậy x = -2

d)(x - \(\frac{2}{9}\)) \(^3\) = (\(\frac{2}{3}\))\(^6\)

(x - \(\frac{2}{9}\)) \(^3\) = \([(\frac{2}{3})^2]^3\)

x - \(\frac{2}{9}\) = \(\frac{4}{9}\)

x = \(\frac{4}{9}\) + \(\frac{2}{9}\)

x = \(\frac{2}{3}\)

Vậy x = \(\frac{2}{3}\)

a) \(\left(\frac{1}{81}\right)^x\cdot27^{2x}=\left(-9\right)^4\)

\(\Leftrightarrow\frac{1}{3^{4x}}\cdot3^{6x}=9^4\)

\(\Leftrightarrow\frac{3^{6x}}{3^{4x}}=3^8\)

\(\Leftrightarrow3^{2x}=3^8\)

\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=4\)

b) \(5^x\cdot\left(5^3\right)^2=625\)

\(\Leftrightarrow5^{x+6}=5^4\)

\(\Leftrightarrow x+6=4\)

\(\Leftrightarrow x=-2\)

c) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

\(\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)

\(\Leftrightarrow\left(4x-1\right)^{20}\cdot\left[\left(4x-1\right)^{10}-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\\left(4x-1\right)^{10}=1=\left(\pm1\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=\frac{1}{2}\\x=0\end{matrix}\right.\)

Vậy....

xin lỡi các bạn nhé

câu a là \(\frac{1}{81}\)