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Cộng vế theo vế
=> \(x^2+x+y^2+y+z^2+z=x^2+y^2+z^2\)
=> \(x+y+z=0\)=> A = 0
\(x=\left(y^2-x^2\right)=\left(y-x\right)\left(y+x\right)=\left(y-x\right).\left(-z\right)=\left(x-y\right).z\)
\(y=\left(z-y\right)\left(z+y\right)=\left(z-y\right).-x=x\left(y-z\right)\)
\(z=y\left(z-x\right)\)
=> \(xyz=\left(x-y\right)\left(y-z\right)\left(z-x\right).xyz\)
=> B = 1
1) \(E^2=\frac{x^2-2xy+y^2}{x^2+2xy+y^2}=\frac{2\left(x^2+y^2\right)-4xy}{2\left(x^2+y^2\right)+4xy}=\frac{5xy-4xy}{5xy+4xy}=\frac{xy}{9xy}=\frac{1}{9}\)
\(\Rightarrow E=\frac{1}{3}\)(vì x>y>0)
2) Ta có \(x+y+z=0\Rightarrow x+y=1-z\)
Lại có : \(1=\left(x+y+z\right)^2=1+2\left(xy+yz+xz\right)\Rightarrow2xy+2yz+2xz=0\Rightarrow2xy=-2z\left(x+y\right)=-2z\left(1-z\right)\)Thay vào \(x^2+y^2+z^2=1\) được :
\(\left(x+y\right)^2-2xy+z^2=1\)\(\Leftrightarrow\left(1-z\right)^2-2z\left(1-z\right)+z^2=1\Leftrightarrow4z^2-4z=0\Leftrightarrow z\left(z-1\right)=0\Leftrightarrow\orbr{\begin{cases}z=0\\z=1\end{cases}}\)
Với z = 0 => x + y = 1 và x2+y2 = 1 => x = 0 , y = 1 hoặc x = 1 , y =0
=> A = 1
Tương tự với z = 1 , ta cũng có x = 0 , y = 0 => A = 1
Bài 2. a/ \(1\le a,b,c\le3\) \(\Rightarrow\left(a-1\right).\left(a-3\right)\le0\) , \(\left(b-1\right)\left(b-3\right)\le0\), \(\left(c-1\right).\left(c-3\right)\le0\)
Cộng theo vế : \(a^2+b^2+c^2\le4a+4b+4c-9\)
\(\Rightarrow a+b+c\ge\frac{a^2+b^2+c^2+9}{4}=7\)
Vậy min E = 7 tại chẳng hạn, x = y = 3, z = 1
b/ Ta có : \(x+2y+z=\left(x+y\right)+\left(y+z\right)\ge2\sqrt{\left(x+y\right)\left(y+z\right)}\)
Tương tự : \(y+2z+x\ge2\sqrt{\left(y+z\right)\left(z+x\right)}\) , \(z+2y+x\ge2\sqrt{\left(z+y\right)\left(y+x\right)}\)
Nhân theo vế : \(\left(x+2y+z\right)\left(y+2z+x\right)\left(z+2y+x\right)\ge8\left(x+y\right)\left(y+z\right)\left(z+x\right)\) hay
\(\left(x+2y+z\right)\left(y+2z+x\right)\left(z+2y+x\right)\ge64\)
Bài 32:
a) P= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(1+\sqrt{2}\)
b) Có: \(x^2-2y^2=xy\)
\(\Leftrightarrow x^2-y^2-y^2-xy=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-y\left(y+x\right)\)
\(\Leftrightarrow\left(x+y\right)\left(x-y-y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y=0\\x-2y=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-y\\x=2y\end{cases}}}\)
Thay x=-y ta có: Q=\(\frac{-y-y}{-y+y}\)=\(\frac{-2y}{0}\)(loại )
Thay x=2y ta có : Q=\(\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)
\(x+y+z=0\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+xz+yz\right)=0\)
\(\Rightarrow1+2\left(xy+xz+yz\right)=0\)
\(\Rightarrow2\left(xy+xz+yz\right)=-1\Rightarrow xy+xz+yz=-\frac{1}{2}\)
\(\Rightarrow\left(xy+xz+yz\right)^2=\frac{1}{4}\)
\(\Rightarrow x^2y^2+x^2z^2+y^2z^2+2xyz\left(x+y+z\right)=\frac{1}{4}\Rightarrow x^2y^2+x^2z^2+y^2z^2=\frac{1}{4}\)
Có:\(\left(x^2+y^2+z^2\right)^2=1\Rightarrow x^4+y^4+z^4+2\left(x^2y^2+x^2z^2+y^2z^2\right)=1\)
\(\Rightarrow x^4+y^4+z^4+\frac{2.1}{4}=1\Rightarrow x^4+y^4+z^4=\frac{1}{2}\)