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

=> 2x + 2 = 2 ( x+ 3  )

=> 2x + 2 = 2x + 6 

=> 2x - 2x = 6 - 2 

=> 0x = 4 ( loại )

Vậy không có số x thỏa mãn 

Ta có:

9^x+3=(3²)^x+3=3^2x+6=3^2x+2

=> 2x+6=2x+2 (vô lý) 

Vậy ko có số x thỏa mãn      

a)

 \(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)

b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)

(2x - 15)⁵ = (2x-15)³

<=> (2x-15)  = 0 hoặc (2x-15) = 1

+ TH1: (2x-15)⁵ = (2x-15)³

             0⁵ = 0³ = 0

+ TH2: (2x-15)⁵ = (2x-15)³

             1⁵ = 1³ = 1

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

\(\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\left(2x-15\right)^3\cdot\left[\left(2x-15\right)^2-1\right]=0\)

\(\hept{\begin{cases}2x-15=0\\\left(2x-15\right)^2-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{15}{2}\\\left(2x-15\right)^2=1\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{15}{2}\\\hept{\begin{cases}2x-15=1\\2x-15=-1\end{cases}\Rightarrow\hept{\begin{cases}x=8\\x=7\end{cases}}}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{15}{2}\\\hept{\begin{cases}2x-15=1\\2x-15=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=8\\x=7\end{cases}}\end{cases}}\)

hello, i'm Trang

( 2x - 15 ) ^5 = ( 2x - 15 ) ^3

=>( 2x - 15 ) ^5 - ( 2x - 15 ) ^3 = 0

=>( 2x - 15 ) ^2 =0

=> 2x-15 = 0

=> x = \(\frac{15}{2}\) =7 \(\frac{1}{2}\)

Vì \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow2x-15=\orbr{\begin{cases}0\\1\end{cases}}\)vì chỉ có \(0^5=0^3;1^5=1^3\)

\(\Rightarrow2x=\orbr{\begin{cases}15\\16\end{cases}}\)

\(\Rightarrow x=\orbr{\begin{cases}\frac{15}{2}\\8\end{cases}}\)

\(\left(2x-15\right)^5=\left(2x-15\right)^3.\)

\(\Rightarrow\hept{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Rightarrow\hept{\begin{cases}2x=15\\2x=16\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{15}{2}\\x=8\end{cases}}}\)

Bài 2

\(a,\left(x-5\right)^4=\left(x-5\right)^6\)

\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)

\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)

\(\Rightarrow\left(x-5\right)^4\left(x-5+1\right)\left(x-5-1\right)=0\)

\(\Rightarrow\left(x-5\right)^4\left(x-4\right)\left(x-6\right)=0\)

\(\Rightarrow x\in\left\{4;5;6\right\}\)

\(b,\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3\left[\left(2xs-15\right)^2-1\right]=0\)

\(\Rightarrow\left(2x-15\right)^3\left(2x-15+1\right)\left(2x-15-1\right)=0\)

\(\Rightarrow\left(2x-15\right)^3\left(2x-14\right)\left(2x-16\right)\)

\(\Rightarrow x\in\left\{\frac{15}{2};7;8\right\}\)

Mà \(\frac{15}{2}\notin n\)

\(\Rightarrow x\in\left\{7;8\right\}\)

#)Giải :

Bài 1 :

a)\(A=\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)

b)\(B=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{11.3^{29}-3^{29}.3}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3^{29}.2^3}{2^2.3^{28}}=6\)

Bài 2 : 

a) \(\left(x-5\right)^2=\left(x-5\right)^6\)

\(\Leftrightarrow x^4-625=x^6-15625\)

\(\Leftrightarrow x^6-x^4=15000\)

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

\(\Leftrightarrow x=5\)

b)\(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Leftrightarrow2x-15=1\)

\(\Leftrightarrow2x=16\)

\(\Leftrightarrow x=8\)

a: \(3x^2\left(x^2-2x+5\right)\)

\(=3x^2\cdot x^2-3x^2\cdot2x+3x^2\cdot5\)

\(=3x^4-6x^3+15x^2\)

b: (x+1)(2x-3)

\(=2x^2-3x+2x-3\)

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

c: \(\left(2x^3-3x^2+x+15\right):\left(2x+3\right)\)

\(=\left(2x^3+3x^2-6x^2-9x+10x+15\right):\left(2x+3\right)\)

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

(2x-  15)⁵ = (2x - 15)³

=> 2x - 15 = 0 ; 2x - 15 = 1 

2x - 15 = 0

2x = 15

x = 7,5

2x - 15 = 1

2x = 16

x = 8        

a) $(15-x)+(x-12)=7-(-8+x)$

$15-x+x-12=7+8-x$

$3=15-x$

$x=12$

Vậy $x=12$.

b) $3(x+2)-6(x-5)=2(5-2x)$

$3x+6-6x+30=10-4x$

$-3x+36=10-4x$

$x=-26$

Vậy $x=-26$.