a: \(\Leftrightarrow2x+7=-4\)

=>2x=-11

hay x=-11/2

b: \(\Leftrightarrow\dfrac{7^x\cdot49+7^x\cdot7+7^x}{57}=\dfrac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)

\(\Leftrightarrow7^x=5^{2x}\)

=>x=0

Ta có: \(\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)

\(\Leftrightarrow\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\)

\(\Leftrightarrow2x+5=4\)

\(\Leftrightarrow2x=-1\)

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

256/625=(4/5)⁴

=>2x+7=4

2x=-3

x=-3/2

d, \(=>\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4.\)

=> \(2x+7=4\) 

=> 2x= -3

=> x=-3/2     . Vậy x=-3/2

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

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

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

=> \(7^x=5^{2x}\)

Đến đoạn này là mik nghĩ không ra nhé

Cô làm tiếp giúp Linh Đan:

\(7^x=5^{2x}\Rightarrow7^x=25^x\Rightarrow\frac{7^x}{25^x}=1\Rightarrow\left(\frac{7}{25}\right)^x=1\Rightarrow x=0\)

( 5/4 ) ^2x+ 7 = ( 5/4 )^4

=> 2x + 7 = 4

=> 2x       = -3

x            = -3/2

xem lại 5/4 hay 4/5

Để sai rồi : 

\(\left(\frac{5}{4}\right)^{2x+7}=\frac{625}{256}\)

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

<=> 2x + 7 = 4

<=> 2x = -3

<=> x = -3/2

Vậy x = -3/2

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

suy ra 2x+7=4

         2x     =-3

          x       = -3/2

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

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

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

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

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

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

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