Question

Tim x biet [x - 2]^{x + 2012}  - [ x - 2]^{x + 2010} = 0

ai giai xong ket ban voi mik nha

Answer 1

\(x^{x+2012}\)-\(2^{x+2012}\)-\(x^{x+2010}\)-\(2^{x+2010}\)=0

x²-2²=0

\(x^2\)-4 =0

x² =0+4=4

=> x=2 hoặc là -2

Answer 2

\(\Leftrightarrow\left(x-2\right)^{x+2010}\left(\left(x-2\right)^2-1\right)=0\)

ĐK :\(x-2\ge1\Leftrightarrow x\ge1\)phuương trình trở thành 

Hoặc : \(\left(x-2\right)^2-1=0\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}TMDK}\)Hoặc : vì theo tính chất lũy thừa nên \(\left(x-2\right)^{x+2010}\ne0\)

KL nghiệm

Answer 3

\(\left(x-2\right)^{x+2012}-\left(x-2\right)^{x+2010}=0\)

\(\Leftrightarrow\left(x-2\right)^{x+2010}\times\left[\left(x-2\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-2\right)^{x+2010}=0\\\left(x-2\right)^2-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-2=1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0+2=2\\x=1+2=3\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)