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\(\dfrac{2011x}{xy+2011x+2011}+\dfrac{y}{yz+y+2011}+\dfrac{z}{xz+z+x}\)
\(=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}\)
\(=\dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+z+1}\)
\(=\dfrac{xz}{1+xz+z}+\dfrac{1}{1+xz+z}+\dfrac{z}{1+xz+z}\)
\(=\dfrac{xz+1+z}{1+xz+z}\)
\(=1\) ( Đpcm )
\(x,y,z\ne0\)
-Ta c/m: -Với \(a+b+c=0\) thì: \(a^3+b^3+c^3-3abc=0\)
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0.\left(a^2+b^2+c^2-ab-bc-ca\right)=0\left(đpcm\right)\)
-Quay lại bài toán:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
\(A=\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=\dfrac{y^3z^3+z^3x^3+x^3y^3}{x^2y^2z^2}=\dfrac{y^3z^3+z^3x^3+x^3y^3-3x^2y^2z^2+3x^2y^2z^2}{x^2y^2z^2}=\dfrac{\left(xy+yz+zx\right)\left[x^2y^2+y^2z^2+z^2x^2-xyz\left(x+y+z\right)\right]}{x^2y^2z^2}+3=\dfrac{0.\left[x^2y^2+y^2z^2+z^2x^2-xyz\left(x+y+z\right)\right]}{x^2y^2z^2}+3=3\)
\(A=\dfrac{x}{xy+x+1}+\dfrac{xy}{x.yz+xy+x}+\dfrac{xy.z}{xy.xz+xy.z+xy}\)
\(=\dfrac{x}{xy+x+1}+\dfrac{xy}{1+xy+x}+\dfrac{1}{x+1+xy}\)
\(=\dfrac{x+xy+1}{xy+x+1}=1\)
Do \(xyz\ne0\) ta có:
\(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}=0\Leftrightarrow xyz\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)=0\Leftrightarrow x+y+z=0\)
Lại có: \(x^3+y^3+z^3=x^3+y^3+3x^2y+3y^2x-3xy\left(x+y\right)+z^3\)
\(=\left(x+y\right)^3+z^3-3xy\left(-z\right)=\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)+3xyz=3xyz\)
Vậy nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz\)
\(P=\dfrac{x^2}{yz}+\dfrac{y^2}{xz}+\dfrac{z^2}{xy}=\dfrac{x^3}{xyz}+\dfrac{y^3}{xyz}+\dfrac{z^3}{xyz}=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{3xyz}{xyz}=3\)
Để M xác định thì \(x,y,z\ne0\)
\(xy+xz+yz=0\Rightarrow\left\{{}\begin{matrix}\dfrac{xy}{z}+x+y=0\\\dfrac{xz}{y}+x+z=0\\\dfrac{yz}{x}+y+z=0\end{matrix}\right.\)
Cộng vế với vế ta được:
\(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}+2\left(x+y+z\right)=0\)
\(\Leftrightarrow M+2.\left(-1\right)=0\Rightarrow M=2\)
Ta có :
\(xy+yz+xz=0\\ \Rightarrow\left[{}\begin{matrix}xy=-xz-yz=-z\left(x+y\right)\\yz=-xy-xz=-x\left(y+z\right)\\xz=-xy-yz=-y\left(x+z\right)\end{matrix}\right.\)
\(M=\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}=\dfrac{-z\left(x+y\right)}{z}+\dfrac{-y\left(x+z\right)}{y}+\dfrac{-x\left(y+z\right)}{x}\\ =-\left(x+y\right)-\left(x+z\right)-\left(y+z\right)=-x-y-x-z-y-z\\ =-2\left(x+y+z\right)=\left(-2\right)\cdot\left(-1\right)=2\)
\(\Rightarrow M=2\)
Các thánh giúp e nha Ace Legona Nguyễn Huy Tú Toshiro Kiyoshi Phương An Akai Haruma @Nguyễn Vũ Phượng Thảo
Ta có: \(\dfrac{1992x}{xy+1992x+1992}\)=
\(\dfrac{xyz.x}{xy+xyz.x+xyz}\) = \(\dfrac{xyz.x.z}{xy.z+xyz.x.z+xyz.z}\) = \(\dfrac{xz}{1+xz+z}\)
Ta có: \(\dfrac{y}{zy+y+1992}\)=\(\dfrac{y}{zy+y+xyz}\)=\(\dfrac{1}{z+1+xz}\)
=> \(\dfrac{1992x}{xy+1992x+1992}\)+\(\dfrac{y}{zy+y+1992}\)+\(\dfrac{z}{z+zx+1}\) = \(\dfrac{xz}{1+zx+z}\) +\(\dfrac{1}{z+zx+1}\) \(+\dfrac{z}{z+zx+1}\) =\(\dfrac{z+zx+1}{z+xz+1}\)
=1