1.
\(3^{100}=\text{(10-1)}^{50}=10^{50}-...+\dfrac{50.49}{2}.10^2-50.10+1\)
\(< =>BS1000+...BS500-500+1=BS1000+1\)
vậy 3^100 có số tận cùng là 001
Còn bài 2 tui chơi nốt
\(x^2+y^2+z^2=x\left(y+z\right)\)
\(\Leftrightarrow x^2+y^2+z^2-xy-xz=0\)
\(\Leftrightarrow4x^2+4y^2+4z^2-4xy-4yz=0\)
\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+\left(4z^2-4yz+y^2\right)+2y^2=0\)
\(\Leftrightarrow\left(2x-y\right)^2+\left(2z-y\right)^2+2y^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2=0\\\left(2z-y\right)^2=0\\2y^2=0\end{matrix}\right.\) \(\Rightarrow x=y=z=0\)
