Question

a, (x-3)^x+8=4*(x-2)^x+6

b, (x-3)^x+10=9*(x-3)^x+8

Answer 1

Đề câu a sai rồi.

`a,(x-3)^{x+8}=4(x-3)^{x+6}`

`=>(x-3)^{x+6}[(x-3)^2-4]=0`

`=>` $\left[ \begin{array}{l}x-3=0\\(x-3)^2=4\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\x-3=2\\x-3=-2\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\x=5\\x=1\end{array} \right.$

Vậy x=1 hoặc x=3 hoặc x=5.

`b,(x-3)^{x+10}=9(x-3)^{x+8}`

`=>(x-3)^{x+8}[(x-3)^2-9]=0`

`=>` $\left[ \begin{array}{l}x-3=0\\(x-3)^2=9\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\x-3=3\\x=-3=-3\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\x=6\\x=0\end{array} \right.$

Vậy x=0 hoặc x=3 hoặc x=6