a. Ta có \(x+2014y+y+2014x=2015\left(x+y\right)⋮2015\)

mà \(\left(x+2014y\right)⋮2015\) \(\Rightarrow\left(y+2014x\right)⋮2015\) ( đpcm )

a. Tớ chưa giải ra được, chờ xíu tớ suy nghĩ :>

b. Ta có \(15x^2-11x-14=\left(5x-7\right)\left(3x+2\right)\)

mà \(15x^2-11x-14⋮7\Rightarrow3x+2⋮7\) ( đpcm )

Thiếu x=-1 nhá

Bữa nay cái đầu cứ bị gì

\(\left(x+1\right)^{x+10}=\left(x+1\right)^{x+4}\)

\(\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\)

\(\Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^{x+6}+1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^{x+4}=0\\\left(x+1\right)^{x+6}-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^{x+6}=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x+1=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

In spite of being interested in arts, he doesn't want to become an artist 

a. Ta có \(A=3+3^2+3^3+...+3^{60}=3\left(1+3+3^2+...+3^{59}\right)⋮3\)

b. Ta có \(A=3+3^2+3^3+...+3^{60}=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)=4\left(3+3^3+...+3^{59}\right)⋮4\)c) Ta có \(A=3+3^2+3^3+...+3^{60}=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)=13\left(3+3^4+...+3^{58}\right)⋮13\)

\(\dfrac{10}{2}=5\), mình đánh nhầm 2 thành 3