a.

\(7^{x+2}+2\times7^{x-1}=345\)

\(7^x\times7^2+2\times7^x\div7=345\)

\(7^x\times\left(7^2+\frac{2}{7}\right)=345\)

\(7^x\times\frac{345}{7}=345\)

\(7^x=345\div\frac{345}{7}\)

\(7^x=345\times\frac{7}{345}\)

\(7^x=7\)

\(x=1\)

b.

\(2^{x+2}-2^x=96\)

\(2^x\times\left(2^2-1\right)=96\)

\(2^x\times3=96\)

\(2^x=\frac{96}{3}\)

\(2^x=32\)

\(2^x=2^5\)

\(x=5\)

\(a,7^{x+2}+2.7^{x-1}=345\Rightarrow7^x.49+\frac{2}{7}.7^x=345\Rightarrow7^x\left(49+\frac{2}{7}\right)=345\Rightarrow7^x.\frac{345}{7}=345\Rightarrow7^x=345:\frac{345}{7}=7^1\Rightarrow x=1\)

\(b,2^{x+2}-2^x=96\Rightarrow2^x.4-2^x=96\Rightarrow2^x\left(4-1\right)=96\Rightarrow2^x.3=96\Rightarrow2^x=96:3=32\Rightarrow2^x=2^5\Rightarrow x=5\)

\(a,7^{x+2}+2.7^{x-1}=345=>7^{x-1+3}+2.7^{x-1}=345=>7^{x-1}.7^3+2.7^{x-1}=345\)

\(=>\left(7^3+2\right).7^{x-1}=345=>345.7^{x-1}=345=>7^{x-1}=1=7^0=>x-1=0=>x=1\)

\(b,2^{x+2}-2^x=96=>2^x.2^2-2^x=96=>2^x.\left(4-1\right)=96=>2^x.3=96=>2^x=32=2^5=>x=5\)

\(\Leftrightarrow7^x\cdot49+2\cdot7^x\cdot\dfrac{1}{7}=345\)

\(\Leftrightarrow7^x=7\)

=>x=1

6) \(\frac{1}{2}.2x+2^{x+2}=2^8+5\)

\(\Rightarrow x+2^{x+2}=2^8+2^5=288\)

- Nếu x < 6 thì x + 2^x+2 < 262

- Nếu x > 6 thì x + 2^x+2 > 519

Vậy không có giá trị nào của x thỏa mãn

b)  \(7^{x+2}+2.7^{x-1}=7^{x-1}.\left(7^3+2\right)=7^{x-1}.345=345\)

\(\Rightarrow7^{x-1}=1\Rightarrow x-1=0\Rightarrow x=1\)

Vậy x = 1 thỏa mãn

tớ nghĩ chắc đề bài ko phải như thế đâu mà là như thế này

7x+2+2.7x−1=3457x+2+2.7x−1=345
Với đề bài này ta làm như sau"
7x+2+2.7x−1=3457x+2+2.7x−1=345

<=>49.7x+27.7x=345<=>49.7x+27.7x=345

<=>3457.7x=345<=>3457.7x=345
<=>7x=7<=>7x=7
<=>x=1<=>x=1

k đi 

k chô mink đi

\(7^x+2+2\cdot7^x-1=345\)

= > 7^x + 2.7^x + ( 2 - 1 )   = 345

=> 7^x .( 2+1) + 1 = 345

=> 7^x .3 = 345 - 1 = 344 

=> 7^x = 344 : 3 = 114,666

\(7^{x-1+3}+2.7^{x-1}=345\Leftrightarrow7^{x-1}\left(7^3+2.1\right)=345\)

\(\Leftrightarrow7^{x-1}.345=345\Leftrightarrow7^{x-1}=1\Leftrightarrow7^{x-1}=7^0\Leftrightarrow x-1=0\Rightarrow x=1\)

Vậy x= 1

=> \(7^{x-1}.7^3+2.7^{x-1}=345\)

=> \(7^{x-1}.\left(7^3+2\right)=345\)

=> \(7^{x-1}.345=345\Rightarrow7^{x-1}=1\Rightarrow x-1=0\) => x = 1 

\(7^{x+2}+2.7^{x-1}=345\)

\(7^{x-1+3}+2.7^{x-1}=345\)

\(7^{x-1}.7^3+2.7^{x-1}=345\)

\(7^{x-1}\left(7^3+2\right)=345\)

\(7^{x-1}\left(343+2\right)=345\)

\(7^{x-1}.345=345\)

\(7^{x-1}=345:345\)

\(7^{x-1}=1\)

\(7^{x-1}=7^0\Rightarrow x-1=0\)

=>\(x=1\)

Vậy x=1

\(7^{x+2}+2.7^{x-1}=345\)

\(\Leftrightarrow7^x.\left(7^2+\dfrac{2}{7}\right)=345\)

\(\Leftrightarrow7x.\dfrac{345}{7}=345\)

\(\Leftrightarrow7x=345:\dfrac{345}{7}\)

\(\Leftrightarrow7x=7\)

\(\Leftrightarrow x=1\)

Vậy x = 1.

a) 2^{x + 2} - 2^x = 96

=> 2^x . 4 - 2^x = 96

=> 2^x . (4 - 1) = 96

=> 2^x . 3 = 96

=> 2^x = 96 : 3

=> 2^x = 32

=> 2^x = 2⁵

=> x = 5

a) 2^{x + 2} - 2^x = 96

=> 2^x. 2² - 2^x = 96

=> 2^x( 4 - 1 ) = 96

=> 2^x = 96 : 3

=> 2^x = 2⁵

=> x = 5

b) 7^{x + 2} + 2 . 7^{x - 1} = 345

=> 7^{x - 1} . ( 7³ + 2 ) = 345

=> 7^{x - 1} . 345 = 345

=> 7^{x - 1} = 345 : 345

=> 7^{x - 1} = 1 = 7⁰

=> x - 1 = 0

=> x = 1

k mik nhé