\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{7^x.\left(7^2+7+1\right)}{57}=7^x\)
\(\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^3}{131}=\frac{5^{2x}\left(1+5+5^3\right)}{131}=\frac{25^x.131}{131}=25^x\)
\(\Rightarrow7^x=25^x\Rightarrow x=0\)
ai tích mình mình tích lại cho
\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{5^{2x}+5^{2x+1}+5^{x+3}}{131}\)
\(\Rightarrow\frac{7^x.49+7^x.7+7^x.1}{57}=\frac{25^x.1+25^x.5+25^x.125}{131}\)
\(\Rightarrow\frac{7^x\left(49+7+1\right)}{57}=\frac{25^x\left(1+5+125\right)}{131}\)
\(\Rightarrow\frac{7^x.57}{57}=\frac{125^x.131}{131}\)
\(\Rightarrow7^x=131^x\)
\(\Rightarrow x=0\)
Vậy \(x=0\)
