Question

5*7^х-1+4*3^х+3^х+1-2*7^х=0(способ группировки)помогите

Answer 1

4*3^x +3^x+1=2*7^x -5*7^x-1)
3^x(4+3)=7^x(2-5*1/7)
3^x*7=7^x*9/7
(3/7)^x=9/49
x=2

Answer 2

$5cdot7 ^{x-1}+4cdot 3 ^{x}+3 ^{x+1}-2cdot7 ^{x}=0, \ 5cdot7 ^{x-1}-2cdot7 ^{x}=-4cdot 3 ^{x}-3 ^{x+1},\7 ^{x-1}(5-2cdot 7)= 3 ^{x}(-4-3), \ 9cdot7 ^{x}cdot frac{1}{7}=7cdot3 ^{x},$
$frac{9}{49}=( frac{3}{7}) ^{x}, \( frac{3}{7}) ^{2}=( frac{3}{7}) ^{x}, \ x=2$