\(7n-1⋮n+2\Leftrightarrow7\left(n+2\right)-8⋮n+2\)

\(\Leftrightarrow-8⋮n+2\Leftrightarrow n+2\inƯ\left(-8\right)=\left\{1;2;4;8\right\}\)

Tự lập bảng 

Tại sao lại ra 7 ( n + 2 ) - 8 vậy bạn

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\)

\(\frac{a+b}{b}=\frac{kb+b}{b}=\frac{b\left(k+1\right)}{b}=k+1\)(1)

\(\frac{c+d}{d}=\frac{kd+d}{d}=\frac{d\left(k+1\right)}{d}=k+1\)(2)

Từ (1) và (2) => đpcm

ta có 

\(4\left(x^2+xy+y^2\right)\ge3\left(x+y\right)^2\Leftrightarrow\left(x-y\right)^2\ge0\) vì thế \(\sqrt{x^2+xy+y^2}\ge\frac{\sqrt{3}}{2}\left(x+y\right)\)

hoàn toàn tương tự ta sẽ có 

\(P\ge\frac{\sqrt{3}}{2}\left(x+y\right)+\frac{\sqrt{3}}{2}\left(y+z\right)+\frac{\sqrt{3}}{2}\left(x+z\right)\)

hay

\(P\ge\sqrt{3}\left(x+y+z\right)=3\sqrt{3}\)

dấu bằng xảy ra khi x=y=z=1

 Ta có: \(P=\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\)\(=\sqrt{\frac{3}{4}\left(x+y\right)^2+\frac{1}{4}\left(x-y\right)^2}+\sqrt{\frac{3}{4}\left(y+z\right)^2+\frac{1}{4}\left(y-z\right)^2}+\sqrt{\frac{3}{4}\left(z+x\right)^2+\frac{1}{4}\left(x-y\right)^2}\)\(\ge\sqrt{\frac{3}{4}\left(x+y\right)^2}+\sqrt{\frac{3}{4}\left(y+z\right)^2}+\sqrt{\frac{3}{4}\left(z+x\right)^2}=\sqrt{3}\left(x+y+z\right)=3\sqrt{3}\)

Đẳng thức xảy ra khi x = y = z = 1

hô hô lêu lêu