Question

[huydeptraichuyengialuadao]

giup minh voi

\(\left[x-1\right]^{x+2}=\left[x-1\right]^{x+4}\)

Answer 1

[huydeptraichuyengialuadao]

ban co the bao minh cach lam ko

Answer 2

[duong]

Ta có: \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)

\(\Leftrightarrow\left(x-1\right)^x.\left(x-1\right)^2=\left(x-1\right)^x.\left(x-1\right)^4\)

\(\Leftrightarrow\left(x-1\right)^2=\left(x-1\right)^4\)

\(\Rightarrow x-1=\left\{-1;1\right\}\)

Nếu x - 1 = - 1 thì x = 0

Nếu x - 1 = 1 thì x = 2

Vậy x mang 2 giá trị là: x =2 hoặc x = 0