mình biến đổi bước xy+yz+zx=3xyz roi nhe 1/x+1/y+1/z=3

Ta có: \(\frac{x^3}{x^2+z}=\frac{x^3+xz}{x^2+z}-\frac{xz}{x^2+z}\ge x-\frac{xz}{2x\sqrt{z}}=x-\frac{\sqrt{z}}{2}\)

Lại có: \(\sqrt{z}\le\frac{z+1}{2}\)

\(\Rightarrow\frac{x^3}{x^2+z}\ge x-\frac{z+1}{4}\)

Tương tự cộng vào ta có: 

\(VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\)

Lại có: \(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)

\(\Rightarrow x+y+z\ge3\)

\(\ge VT\ge\frac{3}{4}.3-\frac{3}{4}=1,5\)

Dấu = xảy ra khi x=y=z=1

Đáp án:

$P=\pm 36$

Giải thích các bước giải:

Ta có:

$\begin{array}{c}{x}^2+y^2+z^2=16\\ xy-yz+zx=-10\\ \Rightarrow \left(x^2+y^2+z^2\right)-2.\left(xy-yz+zx\right)=16-2.\left(-10\right)\\ \iff x^2+y^2+z^2-2xy+2yz-2zx=36\\ \iff \left(x^2-2xy+y^2\right)+z^2+2yz-2zx=36\\ \iff {\left(x-y\right)}^2+2z\left(y-x\right)+z^2=36\\ \iff {\left(x-y\right)}^2-2.\left(x-y\right).z+z^2=36\\ \iff {\left(x-y-z\right)}^2=36\\ \iff x-y-z=\pm 6\\ P=x^3-y^3-z^3-3xyz\\ =\left(x^3-3x^{2}y+3x{y}^2-y^3\right)-z^3+3x^{2}y-3x{y}^2-3xyz\\ ={\left(x-y\right)}^3-z^3+3x^{2}y-3x{y}^2-3xyz\\ =\left[\left(x-y\right)-z\right].\left[{\left(x-y\right)}^2+\left(x-y\right).z+z^2\right]+3xy\left(x-y-z\right)\\ =\left(x-y-z\right).\left(x^2-2xy+y^2+xz-yz+z^2+3xy\right)\\ =\left(x-y-z\right).\left(x^2+y^2+z^2+xy-yz+zx\right)\\ \text{Trường hợp 1:}x-y-z=6\\ \Rightarrow P=6.\left(16+\left(-10\right)\right)=36\\ \text{Trường hợp 2:}x-y-z=-6\\ \Rightarrow P=\left(-6\right).\left(16+\left(-10\right)\right)=-36\end{array}$

Vậy $P=\pm 36$.

MÌNH CHỈ BIẾT LÀM B7 THÔI NHA

P= 811^3+ 812^3+815^3+3.811.812.(-815)=  31694

K ĐÚNG HỘ TỚ NHA

Ta có: 

\(x^3+y^3+z^3-3xyz\)

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

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

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

\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yx-3xz-3yz-3xy\right)\)

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

=> \(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz\)

\(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2zx\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2zx+z^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\x-z=0\end{matrix}\right.\) \(\Leftrightarrow x=y=z\)

Mà \(x+y+z=-3\Rightarrow x=y=z=-1\)

\(\Rightarrow x^2+y^3+z^4=\left(-1\right)^2+\left(-1\right)^3+\left(-1\right)^4=1\)