\(x^2-y^2+z^2-t^2-2xz+2yt=\)

\(=\left(x^2-2xz+z^2\right)-\left(y^2-2yt+t^2\right)=\)

\(=\left(x-z\right)^2-\left(y-t\right)^2=\)

\(=\left[\left(x-z\right)-\left(y-t\right)\right]\left[\left(x-z\right)+\left(y-t\right)\right]\)

   \(x^2-y^2+z^2-t^2-2xz+2yt\)

\(=\left(x^2-2xz+z^2\right)-\left(y^2+2yt+t^2\right)\)

\(=\left(x-z\right)^2-\left(y-t\right)^2\) 

\(=\left(x-z+y-t\right)\times\left(x-z-y+t\right)\)

\(x^2-y^2+z^2-t^2-2xz+2yt\)

\(=x^2-2xz+z^2-\left(y^2-2yt+t^2\right)\)

\(=\left(x-z\right)^2-\left(y-t\right)^2\)

\(=\left(x-z+y-t\right)\left(x-z-y+t\right)\)

a)  \(x^2-8x+16-y^2=\left(x-4\right)^2-y^2=\left(x-y-4\right)\left(x+y-4\right)\)

b)  \(4x^2-y^2+10y-25=4x^2-\left(y-5\right)^2=\left(2x-y+5\right)\left(2x+y-5\right)\)

c)  \(x^2-y^2+z^2-t^2-2xz+2yt=\left(x-z\right)^2-\left(y-t\right)^2=\left(x-y-z+t\right)\left(x+y-x-t\right)\)

p/s: chúc bạn học tốt

\(P=\dfrac{1}{2}\left(2x+4y+6z\right)\left(6x+3y+2z\right)\le\dfrac{1}{8}\left(2x+4y+6z+6x+3y+2z\right)^2\)

\(P\le\dfrac{1}{8}\left(8x+7y+8z\right)^2\le\dfrac{1}{8}\left(8x+8y+8z\right)^2=8\)

\(P_{max}=8\) khi \(\left\{{}\begin{matrix}x+y+z=1\\7y=8y\\2x+4y+6z=6x+3y+2z\end{matrix}\right.\) \(\Leftrightarrow\left(x;y;z\right)=\left(\dfrac{1}{2};0;\dfrac{1}{2}\right)\)

Áp dụng BĐT Svac ta có:
\(P=\dfrac{x^2}{y+3z}+\dfrac{y^2}{z+3x}+\dfrac{z^2}{x+3y}\ge\dfrac{\left(x+y+z\right)^2}{4\left(x+y+z\right)}=\dfrac{x+y+z}{4}=\dfrac{3}{4}\)

Dấu '=' xảy ra khi \(x=y=z=1\)

Vậy \(P_{min}=\dfrac{3}{4}\) khi \(x=y=z=1\)