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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 : \(xy+yz+xz=0\Rightarrow\dfrac{xy+yz+xz}{xyz}=0\)
\(\Leftrightarrow\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}=0\Rightarrow\dfrac{1}{z}=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\dfrac{1}{z^3}=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3\)
\(\Rightarrow\dfrac{1}{z^3}=-\left(\dfrac{1}{x^3}+3.\dfrac{1}{x^2}.\dfrac{1}{y}+3.\dfrac{1}{x}.\dfrac{1}{y^2}+\dfrac{1}{y^3}\right)\)
\(\Rightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3.\dfrac{1}{x}.\dfrac{1}{y}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=3.\dfrac{1}{xyz}\)
Do đó : \(xyz.\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=3\)
\(\Leftrightarrow\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=3\)
\(\Leftrightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
Vậy giá trị của biểu thức \(\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
\(\left\{{}\begin{matrix}xy+yz+xz=0\\x,y,z\ne0\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{x}=0\)\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{y^3}+\dfrac{1}{x^3}=\dfrac{3}{zyz}\)
\(A=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=\dfrac{3xyz}{xyz}=3\)
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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 
Áp dụng công thức a3+b3+c3=3abc
Bài làm
Đặt \(\dfrac{1}{x}\)= a, \(\dfrac{1}{y}\)= b, \(\dfrac{1}{z}\)= c
vì a+b+c = 0 nên a3+b3+c3=3abc
S= \(\dfrac{yz}{x^2}\)+ \(\dfrac{xz}{y^2}\)+ \(\dfrac{xy}{z^{ }2}\)
=\(\dfrac{xyz}{x^{ }3}\)+\(\dfrac{xyz}{y^{ }3}\)+\(\dfrac{xyz}{z^{ }3}\) = xyz(\(\dfrac{1}{x^3}\)+\(\dfrac{1}{y^{ }3}\)+\(\dfrac{1}{z^{ }3}\))
= xyz ( a3+b3+c3 )
= xyz \(\times\)3abc = xyz \(\times\) \(\dfrac{3}{xyz}\) = 3
Ta có:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=-\dfrac{1}{z}\)
\(\Leftrightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\left(-\dfrac{1}{z}\right)^3\)
\(\Leftrightarrow\dfrac{1}{x^3}+3\dfrac{1}{x^2}\dfrac{1}{y}+3\dfrac{1}{x}\dfrac{1}{y^2}+\dfrac{1}{y^3}=-\dfrac{1}{z^3}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}+3\dfrac{1}{x}\dfrac{1}{y}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=0\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}+3\dfrac{1}{x}\dfrac{1}{y}.\left(-\dfrac{1}{z}\right)=0\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)
\(\Leftrightarrow xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=\dfrac{3}{xyz}.xyz\)
\(\Leftrightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
Vậy...
CM bài toán:
nếu a+b+c=0 thì a^3+b^3+c^3=3abc
a^3+b^3+c^3=3abc
=>a^3+b^3+c^3-3abc=0
=>(a+b)^3-3ab(a+b)+c^3-3abc=0
=>[(a+b)^3+c^3]-3ab(a+b+c)=0
=>(a+b+c)[(a+b)^2-(a+b)c+c^2] -3ab(a+b+c)=0
=>(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]=0
vì a+b+c=0 nên a^3+b^3+c^3=3abc
thay a =1/x,b=1/y,c=1/z
áp dụng vào coog thức vừa chứng minh ta đc
\(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)
lại có: A=\(\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz.\dfrac{3}{xyz}=3\)
vậy................
năm mới vui vẻ
\(x,y,z\ne0\)
-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:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
\(A=\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=\dfrac{y^3z^3+z^3x^3+x^3y^3}{x^2y^2z^2}=\dfrac{y^3z^3+z^3x^3+x^3y^3-3x^2y^2z^2+3x^2y^2z^2}{x^2y^2z^2}=\dfrac{\left(xy+yz+zx\right)\left[x^2y^2+y^2z^2+z^2x^2-xyz\left(x+y+z\right)\right]}{x^2y^2z^2}+3=\dfrac{0.\left[x^2y^2+y^2z^2+z^2x^2-xyz\left(x+y+z\right)\right]}{x^2y^2z^2}+3=3\)
\(A=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)}{xyz}\)
\(=\dfrac{\left(-z\right)^3+z^3-3xy\left(-z\right)}{xyz}=3\)