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Xét hạng tử: \(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\)
Thay \(xy+yz+zx=1\); ta có:
\(x\sqrt{\frac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{x^2+xy+yz+zx}}=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)^2\left(x+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}=xy+xz\)
Tượng tự: \(y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}=xy+yz;z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=xz+yz\)
Do đó: \(A=2\left(xy+yz+zx\right)=2.1=2\)
ĐS:...
Bài này hình như x,y,z>0
Ta có: \(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=x\sqrt{\frac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{\left(x^2+xy+yz+zx\right)}}=x\sqrt{\frac{\left(y+x\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}=x\sqrt{\left(y+z\right)^2}\)
Tương tự: \(y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}=y\sqrt{\left(x+z\right)^2}\)
\(z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=z\sqrt{\left(x+y\right)^2}\)
Cộng từng vế, ta có:
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)\)
\(\Leftrightarrow A=2\left(xy+yz+zx\right)=2\)
\(\hept{\begin{cases}1+y^2=y^2+xy+yz+zx=\left(x+y\right)\left(y+z\right)\\1+z^2=\left(z+x\right).\left(z+y\right)\\1+x^2=\left(x+y\right)\left(x+z\right)\end{cases}}\)
Thế vào \(A=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
\(=x\left|y+z\right|+y\left|x+z\right|+z\left|x+y\right|\)
\(=2\left(\left|xy\right|+\left|yz\right|+\left|zx\right|\right)\)
Nếu x,y,z\(\ge0\Rightarrow A=2\)
Nếu x,y,z\(< 0\)\(\Rightarrow A=-2\)
Lời giải:
\(x^3+y^3+8=6xy\)
\(\Leftrightarrow (x+y)^3-3xy(x+y)+8-6xy=0\)
\(\Leftrightarrow [(x+y)^3+2^3]-3xy(x+y+2)=0\)
\(\Leftrightarrow (x+y+2)[(x+y)^2-2(x+y)+4]-3xy(x+y+2)=0\)
\(\Leftrightarrow (x+y+2)(x^2+y^2+4-xy-2x-2y)=0\)
\(\Rightarrow \left[\begin{matrix} x+y+2=0\\ x^2+y^2+4-xy-2x-2y=0\end{matrix}\right.\)
Nếu $x+y+2=0\Rightarrow x+y=-2$
\(P=4(x+y)-(x+2).\frac{(2+y)}{y}.\frac{y+x}{x}=4(x+y)-\frac{(xy+2x+2y+4)(x+y)}{xy}\)
\(=4(-2)-\frac{[xy+2(-2)+4](-2)}{xy}=-8-(-2)=-6\)
Nếu \(x^2+y^2+4-xy-2x-2y=0\)
\(\Leftrightarrow 2x^2+2y^2+8-2xy-4x-4y=0\)
\(\Leftrightarrow (x^2-2xy+y^2)+(x^2-4x+4)+(y^2-4y+4)=0\)
\(\Leftrightarrow (x-y)^2+(x-2)^2+(y-2)^2=0\)
Từ đây dễ dàng suy ra \((x-y)^2=(x-2)^2=(y-2)^2=0\Rightarrow x=y=2\)
\(P=4(2+2)-(2+2)(\frac{2}{2}+1)(\frac{2}{2}+1)=0\)