a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)

=> 2x + 7 = 4 

     2x        = 4 - 7 

     2x        = -3

       x        = -3 : 2

       x         = -1,5

   Vậy x = -1,5

\(\frac{7^{x+2}+7^{x+1}+7x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)

\(\Rightarrow\frac{7x\left(7^2+7^1+1\right)}{57}=\frac{5^{2x}\left(1+5^1+5^3\right)}{131}\)

\(\Rightarrow\frac{7x\left(49+7+1\right)}{57}=\frac{5^{2x}\left(1+5+125\right)}{131}\)

\(\Rightarrow\frac{7x.57}{57}=\frac{5^{2x}.131}{131}\)

\(\Rightarrow7x=25x\)

\(\Rightarrow x=0\)

\(\left(4x-3\right)^4=\left(4x-3\right)^2\)

\(\Rightarrow\left(4x-3\right)^4-\left(4x-3\right)^2=0\)

\(\Rightarrow\left(4x-3\right)^2\left[\left(4x-3\right)^2-1\right]=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(4x-3\right)^2=0\\\left(4x-3\right)^2=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}4x-3=0\\4x-3=-1\\4x-3=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\\x=1\end{cases}}\)

Bài mình làm sai à hay sao mà t]ick sai thế???

a) \(\left(x-4\right)^2=\left(x-4\right)^4\)

\(\Rightarrow\left(x-4\right)^2-\left(x-4^4\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}\left(x-4\right)^2=0\\1-\left(x-4\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\\left(x-4\right)^2=1^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-4=1\\x-4=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=5\\x=3\end{matrix}\right.\)

=>\(\frac{7^x.\left(7^2+7+1\right)}{57}=\frac{5^{2x}.\left(1+5+5^2\right)}{131}\)

=>\(\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)

=>7^x=5^2x

=>7^x=(5²)^x

=>7^x=25^x

=>7=25 (vô lí)

Vậy ko tìm được xthỏa mãn đề bài

x=0 đc mà bạn

\(\frac{7^x\left(7^2+7+1\right)}{57}=\frac{5^{2x}\left(1+5+5^3\right)}{131}\)

\(\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)

\(7^x=5^{2x}\)khi và chỉ khi x = 0.

\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{7^x.\left(7^2+7+1\right)}{57}=7^x\)

\(\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^3}{131}=\frac{5^{2x}\left(1+5+5^3\right)}{131}=\frac{25^x.131}{131}=25^x\)

\(\Rightarrow7^x=25^x\Rightarrow x=0\)

ai tích mình mình tích lại cho

ai tích mình mình tích lại cho