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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 \(x^3+y^3+z^3=3xyz\)
\(P=\dfrac{x^2}{yz}+\dfrac{y^2}{xz}+\dfrac{z^2}{xy}=\dfrac{x^3}{xyz}+\dfrac{y^3}{xyz}+\dfrac{z^3}{xyz}=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{3xyz}{xyz}=3\)
Ta có : \(xy+yz+xz=0\)
\(\Leftrightarrow\dfrac{xy+yz+xz}{xyz}=0\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
C/m 1 bài toán phụ
Cho \(a+b+c=0\) . CM : \(a^3+b^3+c^3=0\)
Do \(a+b+c=0\Rightarrow a+b=-c\Rightarrow\left(a+b\right)^3=-c^3\)
Lại có : \(a^3+b^3+c^3=\left(a+b\right)^3-3ab\left(a+b\right)+c^3=-c^3-3ab\left(-c\right)+c^3=3abc\)
Từ bài toán phụ trên mà ta lại có : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Rightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)
Ta lại có : \(M=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz.\dfrac{3}{xyz}=3\)
Vậy \(M=3\)
Học tốt nhé bạn 
\(VT=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{1}{2}\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{2}\left(\dfrac{1}{z}+\dfrac{1}{x}\right)\ge\dfrac{1}{2}.\dfrac{2}{\sqrt{xy}}+\dfrac{1}{2}.\dfrac{1}{\sqrt{yz}}+\dfrac{1}{2}.\dfrac{1}{\sqrt{zx}}=\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\)
Dấu "=" xảy ra <=> x = y = z > 0
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\dfrac{xy}{z}+\dfrac{yz}{x}\) ≥ \(2\sqrt{\dfrac{xy}{z}.\dfrac{yz}{x}}=2\sqrt{y^2}=2y\left(1\right)\)
\(\dfrac{yz}{x}+\dfrac{xz}{y}\) ≥ \(2\sqrt{\dfrac{yz}{x}.\dfrac{xz}{y}}=2\sqrt{z^2}=2z\left(2\right)\)
\(\dfrac{xy}{z}+\dfrac{xz}{y}\) ≥ \(2\sqrt{\dfrac{xy}{z}.\dfrac{xz}{y}}=2\sqrt{x^2}=2x\left(3\right)\)
Cộng từng vế của ( 1 ; 2 ; 3) , ta được :
\(2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\right)\) ≥ \(2\left(x+y+z\right)\)
⇔ \(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\) ≥ \(x+y+z=2019\)
⇒ \(P_{Min}=2019\) ⇔ \(x=y=z=673\)
\(M=\dfrac{xy+2x+1}{xy+x+y+1}+\dfrac{yz+2y+1}{yz+y+z+1}+\dfrac{xz+2z+1}{xz+z+x+1}\)
\(M=\dfrac{xy+x+x+1}{x\left(y+1\right)+y+1}+\dfrac{yz+y+y+1}{y\left(z+1\right)+z+1}+\dfrac{xz+z+z+1}{z\left(x+1\right)+x+1}\)
\(\Rightarrow M=\dfrac{x\left(y+1\right)+x+1}{\left(x+1\right)\left(y+1\right)}+\dfrac{y\left(z+1\right)+y+1}{\left(y+1\right)\left(z+1\right)}+\dfrac{z\left(x+1\right)+z+1}{\left(z+1\right)\left(x+1\right)}\)
Quy đồng là xong nha
Các thánh giúp e nha Ace Legona Nguyễn Huy Tú Toshiro Kiyoshi Phương An Akai Haruma @Nguyễn Vũ Phượng Thảo
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 :
\(xy+yz+xz=0\\ \Rightarrow\left[{}\begin{matrix}xy=-xz-yz=-z\left(x+y\right)\\yz=-xy-xz=-x\left(y+z\right)\\xz=-xy-yz=-y\left(x+z\right)\end{matrix}\right.\)
\(M=\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}=\dfrac{-z\left(x+y\right)}{z}+\dfrac{-y\left(x+z\right)}{y}+\dfrac{-x\left(y+z\right)}{x}\\ =-\left(x+y\right)-\left(x+z\right)-\left(y+z\right)=-x-y-x-z-y-z\\ =-2\left(x+y+z\right)=\left(-2\right)\cdot\left(-1\right)=2\)
\(\Rightarrow M=2\)