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

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:

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

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

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

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

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

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

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

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

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

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

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

Vậy \(A=3\)

\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\Rightarrow\frac{xy}{ab}=\frac{yz}{bc}=\frac{xz}{ac}=\frac{xy+yz+xz}{ab+bc+ac}.\)(1)

Ta có

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

\(\Leftrightarrow1=1+2\left(ab+bc+ac\right)\Rightarrow ab+bc+ac=0\) => (1) vô nghĩa bạn xem lại đề bài

Lời giải:

Ta có:

$(a+b+c)^2-(a^2+b^2+c^2)=1-1=0$

$\Leftrightarrow 2(ab+bc+ac)=0$

$\Leftrightarrow ab+bc+ac=0$

Đặt $\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=t\Rightarrow x=\frac{a}{t}, y=\frac{b}{t}, z=\frac{c}{t}$

Do đó:

$xy+yz+xz=\frac{ab}{t^2}+\frac{bc}{t^2}+\frac{ac}{t^2}$

$=\frac{1}{t^2}(ab+bc+ac)=\frac{1}{t^2}.0=0$

Ta có 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:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)

\(A=\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=\dfrac{y^3z^3+z^3x^3+x^3y^3}{x^2y^2z^2}=\dfrac{y^3z^3+z^3x^3+x^3y^3-3x^2y^2z^2+3x^2y^2z^2}{x^2y^2z^2}=\dfrac{\left(xy+yz+zx\right)\left[x^2y^2+y^2z^2+z^2x^2-xyz\left(x+y+z\right)\right]}{x^2y^2z^2}+3=\dfrac{0.\left[x^2y^2+y^2z^2+z^2x^2-xyz\left(x+y+z\right)\right]}{x^2y^2z^2}+3=3\)

ta có: \(x+y+z=a\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=a^2\)

\(\Rightarrow b+2\left(xy+yz+xz\right)=a^2\Rightarrow xy+yz+xz=\frac{a^2-b}{2}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{c}\Rightarrow c\left(xy+yz+xz\right)=xyz\)

Ta có:\(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)

\(=a\left(b-\frac{a^2-b}{2}\right)+\frac{3c\left(a^2-b\right)}{2}\)