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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
Ta có: x2+y2+z2=xy+yz+zx (gt)
\(\Leftrightarrow\)2x2+2y2+2z2=2xy+2yz+2zx
\(\Leftrightarrow\)x2-2xy+y2+y2-2yz+z2+z2-2zx+x2=0
\(\Leftrightarrow\)(x-y)2+(y-z)2+(z-x)2=0
\(\Leftrightarrow\)x=y,y=z,z=x
\(\Leftrightarrow\)x=y=z
Khi đó:x2016+y2016+z2016=32017
\(\Leftrightarrow\)3.x2016=32017
\(\Leftrightarrow\)x2016=32016
\(\Leftrightarrow\)x=\(\pm\)3
Vậy:x=y=z=3 hoặc x=y=z=-3
Ta có : \(x^2+y^2+z^2=xy+yz+xz\Leftrightarrow x^2+y^2+z^2-xy-yz-xz=0\)
\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
\(\Leftrightarrow x=y=z\)
Mà \(x^{2016}+y^{2016}+z^{2016}=3^{2017}\)
\(x^{2016}=y^{2016}=z^{2016}=\frac{3^{2017}}{3}=3^{2016}\)
\(\Rightarrow x=y=z=\sqrt[2016]{3^{2016}}=3\)
Ta có:
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]\)
\(=\left(x+y+z\right)^3-3\left(x+y+z\right)\left(x+y\right).z-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yx-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
=> \(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz\)
\(\frac{x^2-yz}{yz}+1+\frac{y^2-zx}{zx}+1+\frac{z^2-xy}{xy}+1=3\Leftrightarrow\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\)
\(\Leftrightarrow\frac{1}{xyz}\left(x^3+y^3+z^3\right)=3\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
Tới đây bạn thay vào nhé :)
\(x^2+y^2+z^2=xy+yz+xz\)
\(2x^2+2y^2+2z^2=2xy+2yz+2xz\)
\(2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)
\(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)=0\)
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
Vì mũ chẵn luôn lớn hơn hoặc bằng 0
\(\Rightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Rightarrow}}x=y=z\)
\(\Rightarrow x^{2015}+y^{2015}+z^{2015}=x^{2015}+x^{2015}+x^{2015}=3x^{2015}\)
\(\Rightarrow3x^{2015}=3^{2016}\)
\(\Rightarrow x^{2015}=3^{2015}\)
\(\Rightarrow x=3\)
Vậy \(x=y=z=3\)
\(M=\frac{x^3+y^3+z^3-3xyz}{x^2+y^2+z^2-xy-yz-zx}\)
Đặt \(N=x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right).z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Vậy \(M=\frac{N}{x^2+y^2+z^2-xy-yz-zx}=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)}{x^2+y^2+z^2-xy-yz-zx}=x+y+z=2016\)
(*) bn ghi sai đề 1 chỗ nhé:ở mẫu thức của M phải là \(x^2+y^2+z^2-xy-yz-zx\) nhé!