CHO \(x+y+z=0\) và \(xy+yz+zx=0\)
TÍNH \(S=\left(x-1\right)^{1999}+y^{2003}+\left(z+1\right)^{2006}\)
Giúp mình vs các bạn
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Bài 1:Áp dụng C-S dạng engel
\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)
Ta có \(x+y+z=0\Leftrightarrow\left(x+y+z\right)^2=0\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)mà xy+yz+zx=0
\(\Rightarrow x^2+y^2+z^2=0\left(1\right)\)
Lại có: \(x^2,y^2,z^2\ge0\Rightarrow x^2+y^2+z^2\ge0\)Kết hợp (1)
\(\Leftrightarrow x^2=y^2=z^2=0\Leftrightarrow x=y=z=0\)
Vậy \(T=\left(0-1\right)^{2013}+0^{2013}+\left(0+1\right)^{2013}=-1+0+1=0\)
Ta có : \(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right)^2=0\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)
\(\Rightarrow x^2+y^2+z^2=0\) ( Do \(xy+yz+zx=0\) )
\(\Rightarrow x^2+y^2+z^2=xy+yz+zx\)
\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)=0\)
\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Rightarrow x=y=z\)
Khi đó : \(x+y+z=3x=0\)
\(\Rightarrow x=0\Rightarrow x=y=z=0\)
Nên \(T=\left(0-1\right)^{2013}+0^{2013}+\left(0+1\right)^{2013}=0\)
Vậy : \(T=0\).
\(\frac{x^2-yz}{yz}+1+\frac{y^2-zx}{zx}+1+\frac{z^2-xy}{xy}+1=3\Leftrightarrow\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\)
\(\Leftrightarrow\frac{1}{xyz}\left(x^3+y^3+z^3\right)=3\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
Tới đây bạn thay vào nhé :)
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)
\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}.\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
\(\ge\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)}.\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)^3}\)
\(\Rightarrow3\left(xy+yz+zx\right)^3\le\left(\dfrac{9}{8}\right)^2\)
\(\Rightarrow\left(xy+yz+zx\right)^3\le\dfrac{27}{64}\)
\(\Rightarrow xy+yz+zx\le\dfrac{3}{4}\)
1111111111111111111
\(VT=\Sigma\frac{xy+yz+zx}{xy}=3+\Sigma\frac{z\left(x+y\right)}{xy}\)
Đến đây để ý \(\frac{1}{2}\left[\frac{z\left(x+y\right)}{xy}+\frac{y\left(z+x\right)}{zx}\right]\ge\sqrt{\frac{\left(z+x\right)\left(x+y\right)}{x^2}}\left(\text{AM - GM}\right)\)
Là xong.
\(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right)^2=0\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
Mà \(xy+yz+xz=0\)
\(\Rightarrow x^2+y^2+z^2+2.0=0\)
\(\Rightarrow x^2+y^2+z^2=0\)
Mà \(x^2\ge0\)
\(y^2\ge0\)
\(z^2\ge0\)
\(\Rightarrow x^2+y^2+z^2\ge0\)
Mà \(x^2+y^2+z^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}\)
\(\Rightarrow B=\left(0-1\right)^{2007}+0^{2008}+\left(0+1\right)^{2009}\)
\(=\left(-1\right)^{2007}+0+1^{2009}\)
\(=-1+0+1\)
\(=0\)
Vậy ...
Do \(x+y+z=0\)
\(\Rightarrow x=-\left(y+z\right)\Rightarrow x^2=\left(y+z\right)^2\Rightarrow4yz-x^2=4yz-\left(y+z^2\right)=-\left(y-z\right)^2\)
Tương tự \(4zx-y^2=-\left(z-x\right)^2\)
\(4xy-z^2=-\left(x-y\right)^2\)
Ta lại có: \(yz+2x^2=yz+x^2-x\left(y+z\right)=yz+x^2-xy-xz=\left(x-y\right)\left(x-z\right)\)
Tương tự: \(zx+2y^2=\left(y-x\right)\left(y-z\right)\)
\(xy+2z^2=\left(y-z\right)\left(y-y\right)\)
\(P=\frac{\left(4yz-x^2\right)\left(4zx-y^2\right)\left(4xy-z^2\right)}{\left(yz+2x^2\right)\left(zx+2y^2\right)\left(xy+2z^2\right)}=\frac{-\left(y-z\right)^2\left(z-x\right)^2\left(x-y^2\right)}{\left(x-y\right)\left(x-z\right)\left(y-x\right)\left(y-z\right)\left(z-x\right)\left(z-y\right)}\)
\(=\frac{-\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2}{-\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2}=1\)
Có xy + yz + zx = 1
=> 1 + x2 = x2 + xy + yz + zx
1 + x2 = (x + y)(y + z)
Tương tự ta có:
1 + y2 = (y + x)(y + z)
1 + z2 = (z + x)(z + y)
Thay vào P, ta được:
\(P=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)
\(P=xy+yz+zx+xy+yz+zx\)
\(P=2\left(xy+yz+zx\right)=2\)
Vậy P = 2
\(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
<=> \(x^2+y^2+z^2=0\Leftrightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}\)
Thay vào tính S:\(S=\left(0-1\right)^{1999}+0^{2003}+\left(0+1\right)^{2006}=-1+1=0\)