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

Sai đề nha bạn

5^x.5^6=5^4

5^x=5^4:5^6

5^x= tap hop rong

Suy ra x= tap hop rong

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

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

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

\(5^x=5^{-2}\)

\(\Rightarrow\text{ }x=-2\)

5^x . ( 5³ )² = 625

<=> 5^x . 5⁶ = 625

<=> 5^x+6 = 625

<=> 5^x+6 = 5⁴

<=> x + 6 = 4

<=> x = -2

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

\(\Leftrightarrow5^x.5^6=625\)

\(\Leftrightarrow5^x=\frac{1}{5^2}\)

\(\Leftrightarrow x=-2\)

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

\(\Rightarrow5^x\times5^6=625\)

      \(5^{x+6}=5^4\)

 \(\Rightarrow x+6=4\)

      \(x=-2\)

    Vậy x = -2

(7.x + 1) : 2 - 23 = 2

=> (7.x + 1) : 2 = 2 + 23

=> (7.x + 1) : 2 = 25

=> 7.x + 1 = 25 . 2

=> 7.x + 1 = 50

=> 7.x = 50 - 1

=> 7.x = 49

=> x = 49 : 7 = 7

(7.x + 1) : 2 - 23 = 2

=> (7.x + 1) : 2 = 2 + 23

=> 7.x + 1 = 25 . 2

=> 7.x = 50 - 1

=> 7.x = 49

 x = 49 : 7 = 7

Vậy.............

hok tốt

x^5=625.x 

<=>x^5:x=625

<=>x^4=625

<=>x^4=5^4

<=>x=5

5^{x .} 5^x+1= 625

5^x+x+1= 625

5^x+x+1=5⁴

=>x+x+1=4

2x+1=4

2x=4-1

2x=3

x=3:2

x=1,5

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) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

                 \(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

                \(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

           \(=>2x+\frac{3}{5}=\frac{3}{5}\)

                                    \(2x=\frac{3}{5}-\frac{3}{5}\)

                                    \(2x=0\)

                                      \(x=0:2\)

                                      \(x=0\)

b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)

=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.

\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

b) \(\left(3x-1\right)\left(-\frac{1}{2}x+5\right)=0\)

\(\left(3x-1\right)\left(-\frac{x}{2}+5\right)=0\)

\(\left(3x-1\right)\left(5-\frac{x}{2}\right)=0\)

\(\orbr{\begin{cases}3x-1=0\\5-\frac{x}{2}=0\end{cases}}\)

\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)