Question

tìm x, biết:

a) (2x-4)^x+1=(2x-4)^x+5

b) (x-3)^x+4=(x-3)^x+7

Answer 1

Có: (2x-4)^x+1=(2x-4)^x+5

<=> (2x-4)^x+1 - (2x-4)^x+5=0

<=> (2x-4)^x+1\([1-\left(2x-4\right)^4]=0\)

<=> \(\left[{}\begin{matrix}\left(2x-4\right)^{x+1}=0\\1-\left(2x-4\right)^4=0\end{matrix}\right.< =>\left[{}\begin{matrix}2x-4=0\\\left(2x-4\right)^4=1\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}2x=4\\2x-4=1hoặc2x-4=-1\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=2\\2x=5hoặc2x=3\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=\dfrac{5}{2};x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy x\(\in\left\{2;\dfrac{5}{2};\dfrac{3}{2}\right\}\)

Answer 2

ý b làm tương tự