a.

 (25)-n. (24)n = 1024
 2-5n. 24n = 210
 2-n = 210
 n = -10

a) \(\left(x-4\right)^2=\left(x-4\right)^4\)

\(\Rightarrow\left(x-4\right)^2-\left(x-4^4\right)=0\)

\(\Rightarrow\left(x-4\right)^2.\left[1-\left(x-4\right)^2\right]=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-4=1\\x-4=-1\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

\(2x+5=4\)

\(2x=-1\)

\(x=-0,5\)

a: \(\Leftrightarrow2x-3=x\)

=>x=3

b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+\dfrac{5}{4}\cdot2^x=\dfrac{7}{32}\)

=>2^x=1/8

=>x=-3

c: =>2x+7=-4

=>2x=-11

=>x=-11/2

d: =>(4x-3)^2*(4x-4)(4x-2)=0

hay \(x\in\left\{\dfrac{3}{4};1;\dfrac{1}{2}\right\}\)

 dề bài đâu mà tìm?

ủa mù à

a: \(5^{x}\cdot\left(5^3\right)^2=625\)

=>\(5^{x}=\frac{5^4}{5^6}=5^{-2}\)

=>x=-2

b: \(\left(\frac{12}{15}\right)^{x}=\left(\frac53\right)^{-5}-\left(-\frac35\right)^4\)

=>\(\left(\frac45\right)^{x}=\left(\frac35\right)^5-\left(\frac35\right)^4=\left(\frac35\right)^4\cdot\left(\frac35-1\right)=\left(\frac35\right)^4\cdot\frac{-2}{5}=\frac{-2\cdot3^4}{5^5}\)

=>\(x=\log_{0,8}\left(-2\cdot\frac{3^4}{5^5}\right)\)

c: \(\left(-\frac34\right)^{3x-1}=\frac{256}{81}\)

=>\(\left(-\frac34\right)^{3x-1}=\left(-\frac34\right)^{-4}\)

=>3x-1=-4

=>3x=-3

=>x=-1

d: \(172x^2-7^9:98^3=2^{-3}\)

=>\(172x^2=\frac18+\frac{7^9}{7^6\cdot2^3}=\frac18+\frac{7^3}{2^3}=\frac{1+343}{8}=\frac{344}{8}\)

=>\(x^2=\frac{344}{8}:172=\frac{344}{8\cdot172}=\frac28=\frac14\)

=>\(\left[\begin{array}{l}x=\frac12\\ x=-\frac12\end{array}\right.\)

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

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

\(\Leftrightarrow2x+5=4\)

\(\Leftrightarrow2x=-1\)

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

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