\(\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\)

dung hằng đẳng thức đẹp :\(x^3+y^3+z^3=3xyz\) với \(x+y+z=0\)

\(\Rightarrow xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz\frac{3}{xyz}=3\)

Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Rightarrow\hept{\begin{cases}1+\frac{x}{y}+\frac{x}{z}=0\\\frac{y}{x}+1+\frac{y}{z}=0\\\frac{z}{x}+\frac{z}{y}+1=0\end{cases}}\)

\(\Rightarrow\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}=-3\)

mà \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Rightarrow\frac{yz+xz+xy}{xyz}=0\)

\(\Rightarrow yz+xz+xy=0\)

\(\Rightarrow\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\left(yz+xz+xy\right)=0\)

\(\Rightarrow\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}=0\)

\(\Rightarrow\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=3\)

\(\Rightarrow\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{xz}{y^2}=3\)

Học tốt

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

<=> \(\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)

<=> \(\left(\frac{1}{x}+\frac{1}{y}\right)^3=\left(-\frac{1}{z}\right)^3\)

<=> \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{x^2y}+\frac{3}{xy^2}=-\frac{1}{z^3}\)

<=> \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=-\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)\)

<=> \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=-\frac{3}{xy}.\left(-\frac{1}{z}\right)\)

<=> \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)

Khi đó: P = \(\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{xz}{y^2}=\frac{xyz}{z^3}+\frac{xyz}{x^3}+\frac{xyz}{y^3}=xyz.\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz\cdot\frac{3}{xyz}=3\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

<=>xy+yz+zx=0 (x,y,z khác 0)

P=\(\frac{\left(xy\right)^2+\left(yz\right)^2+\left(yz\right)^2}{\left(xyz\right)^2}\)

=\(\frac{\left(xy+yz+zx\right)^2}{\left(xyz\right)^2}\)-\(\frac{2xyz\left(x+y+z\right)}{\left(xyz\right)^2}\)

=\(-\frac{2x+2y+2z}{xyz}\)

Với x,y,z khác 0 ta có \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0=>\frac{yz+xz+xy}{xyz}=0=>yz+xz+xy=0\)

Ta luôn có nếu a+b+c=0 thì a³+b³+c³=3abc

Vì xy+yz+zx=0 nên x³y³+y³z³+z³x³=3x²y²z²

Với x³y³+y³z³+z³x³=3x²y²z² ta có:

\(\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{y^3z^3+x^3z^3+x^3y^3}{x^2y^2z^2}=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)

Vậy ....

Ta có:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

\(\Rightarrow\dfrac{1}{z}=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Rightarrow\left(\dfrac{1}{z}\right)^3=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3\)

\(\Rightarrow\dfrac{1}{z^3}=-\left(\dfrac{1}{x^3}+3\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}+3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y^2}+\dfrac{1}{y^3}\right)\)

\(\Rightarrow\dfrac{1}{z^3}=-\dfrac{1}{x^3}-\dfrac{3}{x^2y}-\dfrac{3}{xy^2}-\dfrac{1}{y^3}\)

\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3}=-3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3}=-3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot-\dfrac{1}{z}\)

\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3}=3\cdot\dfrac{1}{xyz}\)

\(\Rightarrow xyz\cdot\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=3\)

\(\Rightarrow\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=3\)

\(\Rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)

Vậy \(A=3\)

nhấn vào đúng 0 sẽ ra kết quả mình làm bài này rồi

lm đi r mk cho

câu này khó thế cậu 

sorry mình không biết câu này

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