Question

Tìm x:

1) \(\left(4x-7\right)^2-5\times\left|7-4x\right|=0\)

2) \(4^{x-2}+4^{x+1}=1040\)

Answer 1

1)

Có (4x-7)² \(\ge0\) với mọi x

|7−4x| \(\ge0\) với mọi x

<=> 5|7−4x| \(\ge0\) với mọi x

Để <=>\(\left\{{}\begin{matrix}4x-7=0\\7-4x=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}4x=7\\4x=7\end{matrix}\right.\)<=>\(x=\dfrac{7}{4}\)

Vậy \(x=\dfrac{7}{4}\)

Answer 2

2) \(4^{x-2}+4^{x+1}=1040\)

<=> \(4^{x+1}.4^{-3}+4^{x+1}=1040\)

<=> \(4^{x+1}\left(4^{-3}+1\right)=1040\)

<=> \(4^{x+1}.\dfrac{65}{64}=1040\)

<=> \(4^{x+1}=1024=4^5\)

=> x+1=5 <=> x=4

Vậy x=4