\(x^{x+2012}\)-\(2^{x+2012}\)-\(x^{x+2010}\)-\(2^{x+2010}\)=0
x2-22=0
\(x^2\)-4 =0
x2 =0+4=4
=> x=2 hoặc là -2
\(\Leftrightarrow\left(x-2\right)^{x+2010}\left(\left(x-2\right)^2-1\right)=0\)
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
\(\left(x-2\right)^{x+2012}-\left(x-2\right)^{x+2010}=0\)
\(\Leftrightarrow\left(x-2\right)^{x+2010}\times\left[\left(x-2\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-2\right)^{x+2010}=0\\\left(x-2\right)^2-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-2=1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0+2=2\\x=1+2=3\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
