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ta có \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz.\)lúc đó
\(P=\frac{2018\left(x-y\right)\left(y-z\right)\left(z-x\right)}{2xy^2+2yz^2+2zx^2+3xyz}=2018.\frac{xy^2+yz^2+zx^2-x^2y-y^2z-z^2x}{xy^2+yz^2+zx^2+y^2\left(x+y\right)+x^2\left(x+z\right)+z^2\left(z+y\right)}\)
\(P=2018.\frac{xy^2+yz^2+zx^2-x^2y-y^2z-z^2x}{xy^2+yz^2+zx^2-x^2y-y^2z-z^2x}=2018\)
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 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\(< 0\)\(\Rightarrow A=-2\)
Bài 1 :
Ta có :
\(x^7+\frac{1}{x^7}=\left(x^3+\frac{1}{x^3}\right)\left(x^4+\frac{1}{x^4}\right)-\left(x+\frac{1}{x}\right)\)
\(\left(x+\frac{1}{x}\right)=a\Leftrightarrow\left(x+\frac{1}{x}\right)^2=a^2\)
\(\Leftrightarrow x^2+\frac{1}{x^2}+2.x.\frac{1}{x}=a^2\)
\(\Leftrightarrow x^2+\frac{1}{x^2}=a^2-2\)
\(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)\left(x^2-x.\frac{1}{x}+\frac{1}{x^2}\right)\)
\(=a\left(x^2+\frac{1}{x^2}-1\right)=a\left(a^2-3\right)\)
\(x^4+\frac{1}{x^4}=\left(x^2+\frac{1}{x^2}\right)^2-2.x^2.\frac{1}{x^2}\)
\(=\left(a^2-2\right)^2-2=a^4-4a^2+4-2\)
\(=a^4-4a^2+2\)
\(\Rightarrow x^7+\frac{1}{x^7}=a.\left(a^2-3\right).\left(a^4-4a^2+2\right)-a\)
\(=\left(a^3-3a\right)\left(a^4-4a^2+2\right)-a\)
\(=a^7-4a^5+2a^3-3a^5+12a^3-6a-a\)
\(=a^7-7a^5+14a^3-7a\)
Bài 2 :
Ta có :
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=2^2\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=4\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=\frac{2}{xy}-\frac{1}{z^2}\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{z^2}+\frac{2}{yz}+\frac{2}{zx}=0\)
\(\Rightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\)
\(\Rightarrow\frac{1}{x}+\frac{1}{z}=\frac{1}{y}+\frac{1}{z}=0\) vì \(\left(\frac{1}{x}+\frac{1}{z}\right)^2,\left(\frac{1}{y}+\frac{1}{z}\right)^2\ge0\)
\(\Rightarrow x=y=-z\)
\(\Rightarrow\frac{1}{-z}+\frac{1}{-z}+\frac{1}{z}=2\Rightarrow-\frac{1}{z}=2\Rightarrow z=-\frac{1}{2}\)
\(\Rightarrow x=y=\frac{1}{2}\)
\(\Rightarrow x+2y+z=\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}=1\)
\(\Rightarrow P=1\)