ko biết

\(x+y+z+t=0\Rightarrow t=-\left(x+y+z\right)\)

Ta có: 

\(VT=x^3+y^3+z^3+t^3=x^3+y^3+z^3-\left(x+y+z\right)^3\)

\(=x^3+y^3+z^3-\left[x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\right]\)

\(=-3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(VP=3\left[xy+z\left(x+y+z\right)\right]\left(z-x-y-z\right)=3\left(xy+yz+zx+z^2\right)\left(-x-y\right)\)

\(=-3\left(y+z\right)\left(z+x\right)\left(x+y\right)\)

Do VT = VP nên ta có đpcm.

Bạn xem lại đề. Nếu không có thêm điều kiện gì của $x,y,z$ (chả hạn nguyên, nguyên dương) thì không giải được đâu.

\(\frac{1}{x+y+z}+\frac{1}{3}=\frac{1}{x+y+z}+\frac{1}{3xyz}\ge\frac{2}{\sqrt{3xyz\left(x+y+z\right)}}\ge\frac{2}{xy+yz+zx}\)

Dấu "=" xảy ra khi \(x=y=z=1\)

Lời giải:

Ta có: \(\left\{\begin{matrix} xy+x+y=3\\ yz+y+z=8\\ zx+z+x=15\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (x+1)(y+1)=4\\ (y+1)(z+1)=9\\ (z+1)(x+1)=16\end{matrix}\right.(1)\)

Nhân 3 vế với nhau:

\(\Rightarrow [(x+1)(y+1)(z+1)]^2=4.9.16\)

\(\Leftrightarrow (x+1)(y+1)(z+1)=\pm 24\)

Nếu \((x+1)(y+1)(z+1)=24(2)\)

Từ \((1),(2)\Rightarrow \left\{\begin{matrix} z+1=6\\ x+1=\frac{8}{3}\\ y+1=\frac{3}{2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=\frac{5}{3}\\ y=\frac{1}{2}\\ z=5\end{matrix}\right.\)

Do đó, \(P=x+y+z=\frac{43}{6}\)

Nếu 

\((x+1)(y+1)(z+1)=-24(3)\)

Từ $(1);(3)$ suy ra \(\left\{\begin{matrix} z+1=-6\\ x+1=\frac{-8}{3}\\ y+1=\frac{-3}{2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} z=-7\\ x=-\frac{11}{3}\\ y=\frac{-5}{2}\end{matrix}\right.\)

Do đó, \(P=x+y+z=-\frac{79}{6}\)

Đề sai.Sửa đề: \(xy+x+y=3\)

\(\left\{{}\begin{matrix}xy+x+y=3\\yz+y+z=8\\xz+x+z=15\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}xy+x+y+1=4\\yz+y+z+1=9\\xz+x+z+1=16\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y\left(x+1\right)+1\left(x+1\right)=4\\y\left(z+1\right)+1\left(z+1\right)=9\\x\left(z+1\right)+1\left(z+1\right)=16\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=4\\\left(y+1\right)\left(z+1\right)=9\\\left(x+1\right)\left(z+1\right)=16\end{matrix}\right.\)

\(\Rightarrow\left(x+1\right)\left(y+1\right)\left(y+1\right)\left(z+1\right)\left(x+1\right)\left(z+1\right)=576\)

\(\Rightarrow\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2=576\)

\(\Rightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)=24\)

Đến đây chịu :v