\(4^{x-5}=16\)

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

\(x-5=2\)

\(x=2+5\)

\(x=7\)

\(45-2^{x-1}=29\)

\(2^{x-1}=16\)

\(2^{x-1}=2^4\)

\(x-1=4\)

\(x=5\)

\(\left(2+x\right)^2=144\)

\(\left(2+x\right)^2=12^2\)

\(2+x=12\)

\(x=12-2\)

\(x=10\)

\(\left(x-5\right)^2=81\)

\(\left(x-5\right)^2=9^2\)

\(x-5=9\)

\(x=14\)

\(\left(13-x\right)^4=81\)

\(\left(13-x\right)^4=3^4\)

\(13-x=3\)

\(x=13-3\)

\(x=10\)

\(...4^{x-5}=4^2\Rightarrow x-5=2\Rightarrow x=7\)

\(...2^{x-1}=45-29=16\Rightarrow2^{x-1}=2^4\Rightarrow x-1=4\Rightarrow x=5\)

\(...\Rightarrow\left(2+x\right)^2=12^2\Rightarrow\left[{}\begin{matrix}2+x=12\\2+x=-12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-14\end{matrix}\right.\)

\(...\Rightarrow\left(x-5\right)^2=9^2\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

\(...\Rightarrow\left(13-x\right)^4=3^4\Rightarrow\left[{}\begin{matrix}13-x=3\\13-x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=16\end{matrix}\right.\)

Mình quên mũ 2 với mũ 4 là chia ra 2 trường hợp, sửa lại nhé:

\(\left(2+x\right)^2=144\)

 \(\left(2+x\right)^2=12^2\)

Th1: 

\(2+x=12\)

\(x=10\)

Th2: 

\(2+x=-12\)

\(x=-14\)

\(\left(x-5\right)^2=81\)

\(\left(x-5\right)^2=9^2\)

Th1: 

\(x-5=9\)

\(x=14\)

Th2: 

\(x-5=-9\)

\(x=-4\)

\(\left(13-x\right)^4=81\)

\(\left(13-x\right)^4=3^4\)

Th1: 

\(13-x=3\)

\(x=10\)

Th2: 

\(13-x=-3\)

\(x=16\)

a) \(3^{x-1}+3x+3^{x+1}=1053\)

\(=3^x:3+3^x+3^x.3=1053\)

\(=3^x.\dfrac{1}{3}+1+3=1053\)

\(=3^x.\dfrac{13}{5}=1053\)

\(=3^x=243\)

\(\Rightarrow x=5\)

Vậy \(x=5\)

a/

x= 918 : 9 = 102

b/

13590 : 45 = 302

c/

x = 1834 . 15 = 27510

d/

6.x = 240

=> x= 240 : 6 = 40

e/

5.x = 450

=> x= 450 : 5 = 90

f/

21-x = 9

=> x= 21-9 = 12

a. => [(2x+14):2²-3]:2=1

=> (2x+14):4-3=2

=> (2x+14):4=5

=> 2x+14=20

=> 2x=20-14

=> 2x=6

=> x=6:2

Vậy x=3.

b. 3^x+1=81²

=> 3^x+1=(3⁴)²

=> 3^x+1=3⁸

=> x+1=8

Vậy x=7.

c. 3^x+3^x+1+3^x+2=1053

=> 3^x.(1+3+3²)=1053

=> 3^x.13=1053

=> 3^x=81

=> 3^x=3⁴

Vậy x=4.

a. 3^x+3^x+1+3^x+2=1053

=> 3^x.(1+3+3²)=1053

=> 3^x.(1+3+9)=1053

=> 3^x.13=1053

=> 3^x=1053:13

=> 3^x=81

=> 3^x=3⁴

=> x=4

b. 5^x.5¹⁹=5²⁰.5¹¹

=> 5^19+x=5²⁰⁺¹¹

=> 19+x=20+11

=> 19+x=31

=> x=31-19

=> x=12

\(3^x+3^{x+1}+3^{x+2}=1053\)

\(3^x\left(1+3+3^2\right)=1053\)

\(3^x\cdot13=1053\)

\(3^x=1053:13=81\)

\(3^x=3^4\)

=> x = 4 

A)x=4

B)x=12 100% đó

3^x+3^(x+1)+3^(x+2)=1053 <=> 3^x+3^1*3^x+3^2*3^x=1053 
<=> (1+3^1+3^2)*3^x=1053 
<=>3^x= 1053/ (1+3+9) 
<=> 3^x=81 
=> x=4

3^x+3^(x+1)+3^(x+2)=1053 <=> 3^x+3^1*3^x+3^2*3^x=1053 

<=> (1+3^1+3^2)*3^x=1053 
<=>3^x= 1053/ (1+3+9) 
<=> 3^x=81 
=> x=4

Đây là 2 phương án. Cái nào đúng đề của bạn ban đầu thì bạn chọn nhé! ( do tớ không hiểu là 3^x + 1 hay là 3 ^{x + 1} )
Phương án 1:
3^x + 3^x + 1 + 3^x + 2 = 1053
=> 3 * 3^x + 3 = 1053
=> 3 ^{1 + x} = 1050
=> 1 + x = \(\log_31050\)
=> 1 + x \(\approx\) 6,3321
=> x \(\approx\) 5,3321
Phương án 2:
3^x + 3^{x + 1} + 3^{x + 2} = 1053
=> 3^x + 3^x * 3 + 3^x + 9 = 1053
=> 3^x ( 1 + 3 + 9 ) = 1053
=> 3^x * 13 = 1053
=> 3^x = 1040
=> x = \(\sqrt[3]{1040}\)
=> x = 4

eo biet