Vì x+y+z=6 và \(x^2+y^2+z^2=12\)

Ta có \(x^2+y^2+z^2-x+y+z=12-6\)

Rút gọn: \(x\left(x-1\right)+y\left(y-1\right)+z\left(z-1\right)=6\)

=> \(x+y+z=x\left(x-1\right)+y\left(y-1\right)+z\left(z-1\right)\)

Tìm x \(\Rightarrow x\left(x-1\right)=x\Rightarrow x-1=1\Rightarrow x=2\)

Tìm y \(\Rightarrow y\left(y-1\right)=y\Rightarrow y-1=1\Rightarrow y=2\)

Tìm z \(\Rightarrow z\left(z-1\right)=z\Rightarrow z-1=1\Rightarrow z=2\)

Vậy \(x=y=z=2\)

\(\hept{\begin{cases}x^2+y^2+z^2=12\\x+y+z=6\end{cases}}\)

Ta có \(\left(x+y+z\right)^2=36\)

\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz=36\)

\(\Leftrightarrow12+2xy+2yz+2xz=36\)

\(\Leftrightarrow2xy+2yz+2xz=24\Leftrightarrow xy+yz+xz=12\)

\(\Rightarrow x^2+y^2+z^2=xy+yz+xz=12\)

Mặt khác ta có \(x^2+y^2+z^2\ge xy+yz+xz\)

Dấu \(=\)xảy ra khi \(x=y=z\)

Vậy \(x=y=z=2\)

Câu trả lời là thiếu dự kiện

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)

Trường hợp x=y=z thì không phải bàn,ns cái trường hợp x+y+z=0

\(\frac{1}{x^2+y^2-z^2}=\frac{1}{\left(x+y\right)^2-2xy-z^2}=\frac{1}{\left(-z\right)^2-z^2-2xy}=\frac{1}{-2xy}\)

Tương tự rồi cộng lại thì \(BT=0\) thì phải

Condition\(\hept{\begin{cases}x\ne0\\y\ne0\\z\ne0\end{cases}}\)

Put \(P=\frac{1}{x^2+y^2-z^2}+\frac{1}{y^2+z^2-x^2}+\frac{1}{z^2+x^2-y^2}\)

\(=\frac{1}{x^2+\left(y-z\right)\left(y+z\right)}+\frac{1}{y^2+\left(z-x\right)\left(z+x\right)}+\frac{1}{z^2+\left(x-y\right)\left(x+y\right)}\left(4\right)\)

Because \(x^2+y^2+z^2=3xyz\)

\(\Leftrightarrow x^2+y^2+z^2-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=0\)ư\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2yz-2zx\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\end{cases}}\)

The first case: If \(x+y+z=0\left(1\right)\)

\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\z+x=-y\end{cases}\left(2\right)}\)

From \(\left(1\right)\Rightarrow\hept{\begin{cases}x-y=-2y-z\\y-z=-2z-x\\z-x=-2x-y\end{cases}\left(3\right)}\)

 \(\left(2\right)\)and \(\left(3\right)\)into \(\left(4\right)\)we have

\(P=\frac{1}{x^2-x\left(-2z-x\right)}+\frac{1}{y^2-y\left(-2x-y\right)}+\frac{1}{z^2-z\left(-2y-z\right)}\)

\(=\frac{1}{2x^2+2xz}+\frac{1}{2y^2+2xy}+\frac{1}{2z^2+2yz}\)

\(=\frac{1}{2x\left(x+z\right)}+\frac{1}{2y\left(x+y\right)}+\frac{1}{2z\left(z+y\right)}\)

\(\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}\)

\(\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}\)

\(=\frac{z+x+y}{-2xyz}=0\)( Because x+y+z=0)

The second case:\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\left(5\right)\)

We have \(\hept{\begin{cases}\left(x-y\right)^2\ge0;\forall x,y,z\\\left(y-z\right)^2\ge0;\forall x,y,z\\\left(z-x\right)^2\ge0;\forall x,y,z\end{cases}}\)\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0;\forall x,y,z\left(6\right)\)

From \(\left(5\right),\left(6\right)\)\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Leftrightarrow x=y=z}\)

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

So \(P=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)

\(=\frac{z+x+y}{xyz}=0\)

So...

Bạn kia làm ra kết quả đúng nhưng cách làm thì tào lao nhưng vẫn ra ???

Áp dụng BĐT Cô-si ta có:

\(\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\ge3\sqrt[3]{\frac{1}{x\left(x+1\right)}.\frac{x}{2}.\frac{x+1}{4}}=\frac{3}{2}\)

Tương tự:\(\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{4}\ge\frac{3}{2}\),\(\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge\frac{3}{2}\)

Cộng vế với vế của 3 BĐT trên ta được:

\(P+\frac{x+y+z}{2}+\frac{\left(x+y+z\right)+3}{4}\ge\frac{9}{2}\)

\(\Leftrightarrow P+\frac{3}{2}+\frac{6}{4}\ge\frac{9}{2}\)

\(\Leftrightarrow P\ge\frac{3}{2}\)

Dấu '=' xảy ra khi \(\hept{\begin{cases}\frac{1}{x^2+x}=\frac{x}{2}=\frac{x+1}{4}\\\frac{1}{y^2+y}=\frac{y}{2}=\frac{y+1}{4}\\\frac{1}{z^2+z}=\frac{z}{2}=\frac{z+1}{4},x+y+z=3\end{cases}\Leftrightarrow x=y=z=1}\)

Vậy \(P_{min}=\frac{3}{2}\)khi \(x=y=z=1\)

Áp dụng bđt Bunhiacopski ta có

\(P\ge\frac{9}{x^2+y^2+z^2+x+y+z}\ge\frac{9}{2\left(x+y+z\right)}=\frac{9}{6}=\frac{3}{2}.\)

Dấu "=" xảy ra khi x=y=z=1

Ta có : \(3\left(x^2+y^2+z^2\right)=\left(x+y+z\right)^2\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)

\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

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

\(\Leftrightarrow x=y=z\)

Khi đó : \(3x^{2018}=27^{673}=\left(3^3\right)^{673}=3^{2019}\)

\(\Leftrightarrow x^{2018}=3^{2018}\)

\(\Leftrightarrow\orbr{\begin{cases}x=y=z=3\\x=y=z=-3\end{cases}}\)

Đến đây tự tính A nha!

áp dụng bất đẳng thức cauchy cho 2015 số , ta có

\(2x^{2015}+2013=x^{2015}+x^{2015}+1+1+..+1\ge2015\sqrt[2015]{x^{2015}.x^{2015}}=2015x^2\)

tương tự ta có

\(\hept{\begin{cases}2.y^{2015}+2013\ge2015y^2\\2.z^{2015}+2013\ge2015z^2\end{cases}}\)

cộng ba bất đẳng thức lại ta có \(2\left(x^{2015}+y^{2015}+z^{2015}\right)+2013.3\ge2015\left(x^2+y^2+z^2\right)\)

hay \(2015\left(x^2+y^2+z^2\right)\le2.3+2013.3=2015.3\Rightarrow\left(x^2+y^2+z^2\right)\le3\)

dấu "=" xảy ra khi x=y=z=1