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

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

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

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

 dề bài đâu mà tìm?

ủa mù à

a) \(5^x-\left(5^3\right)^2=625\)

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

\(\Leftrightarrow x-6=4\)

\(\Leftrightarrow x=10\)

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

\(5^x.\left(5^3\right)^2=625=5^4\)

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

Vậy x = -2 

\(\left(\frac{3}{5}\right)^x.\left(\frac{625}{81}\right)^3=\frac{243}{3125}\)

=>\(\left(\frac{3}{5}\right)^x.\left(\frac{5}{3}\right)^{12}=\left(\frac{3}{5}\right)^5\)

=>\(\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^5:\left(\frac{5}{3}\right)^{12}\)

=>\(\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^{17}\)

=>x=17

\(\left(\frac{3}{5}\right)^x.\left(\frac{625}{81}\right)^3=\frac{243}{3125}\)

\(\Rightarrow\left(\frac{3}{5}\right)^x.\left(\frac{3}{5}\right)^{12}=\left(\frac{3}{5}\right)^5\)

\(\Rightarrow\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^5:\left(\frac{3}{5}\right)^{12}\)

\(\Rightarrow\left(\frac{3}{5}\right)^x=\left(\frac{5}{3}\right)^7\)

\(\Rightarrow x=-7\)

Bài làm:

a) Ta có: \(5^{x+2}=625\)

\(\Leftrightarrow5^{x+2}=5^4\)

\(\Rightarrow x+2=4\)

\(\Rightarrow x=2\)

b) \(\left(x-1\right)^{x+2}=\left(-1\right)^{x+4}\)

\(\Leftrightarrow\left(x-1\right)^{x+2}=\left(-1\right)^{x+2}.\left(-1\right)^2\)

\(\Leftrightarrow\left(x-1\right)^{x+2}=\left(-1\right)^{x+2}\)

\(\Rightarrow x-1=-1\)

\(\Rightarrow x=0\)

c) \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Rightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\)

\(\Rightarrow x=-\frac{1}{2}\)

5^x+2=625

5^x+2=5^4

x+2=4

x=4-2

x=2

(x-1)^x+2=[(-1)^2]^x+2

(x-1)=(-1)^2

(x-1)=1

x=1+1

x=2

vậy x=2

(2x-1)^3=-8

(2x-1)^3=(-2)^3

2x-1=-2

2x=-2+1

2x=-1

x=-1:2

x=-0,5

vậy x=-0,5

vậy x=2

Ta có: 25^{x + 1} . 125^x . 625^{x + 2} = (5²)⁵

=> (5²)^{x + 1} . (5³)^x  . (5⁴)^{x+ 1} = 5¹⁰

=> 5^{2x + 2} . 5^3x . 5^{4x + 8} = 5¹⁰

=> 2x + 2 + 3x + 4x + 8 = 10

=> 9x + 2 + 8 = 10

=> 9x = 10 - 2 - 8

=> 9x = 0

=> x = 0 : 9

=> x = 0

a.\(3x^2-51=-24\)

\(\Rightarrow3x^2=27\)

\(\Rightarrow x^2=9\)

\(\Rightarrow x=3\)

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

\(\Rightarrow5x=5^5:5^{\left(3x2\right)}\)

\(\Rightarrow5x=5^5:5^6\)

\(\Rightarrow5x=5^{-1}=0,2\)

\(\Rightarrow x=0,2:5\)

\(\Rightarrow x=0,04\)

Bài 1:

Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)

\(\Leftrightarrow2x=\frac{1440}{144}=10\)

\(\Rightarrow x=5\)

Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)

=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)