lấy mẫu trừ đi (ax+by+cz)^2

Áp dụng : A = - A => A = 0

Từ \(a+b+c=0\Rightarrow\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(c+a\right)\\c=-\left(a+b\right)\end{cases}}\)

  \(x+y+z=0\Rightarrow\hept{\begin{cases}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(x+y\right)\end{cases}\Rightarrow\hept{\begin{cases}x^2=\left(y+z\right)^2\\y^2=\left(x+z\right)^2\\z^2=\left(x+y\right)^2\end{cases}}}\)

Và \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=\frac{ayz+bxz+cxy}{xyz}=0\)\(\Rightarrow ayz+bxz+cxy=0\)

Ta có : \(x^2a+y^2b+x^2c=\)\(\left(y+z\right)^2a+\left(x+z\right)^2b+\left(x+y\right)^2c\) 

=   \(x^2\left(b+c\right)+y^2\left(c+a\right)+z^2\left(a+b\right)\)\(+2\left(ayz+bxz+cxy\right)\)

=  \(-\left(x^2a+y^2b+z^2c\right)\) => \(x^2a+y^2b+x^2c=\) 0 

Ta có: \(\hept{\begin{cases}a=-b-c\\x=-y-z\end{cases}}\)

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)

\(\Leftrightarrow\frac{\left(-b-c\right)}{\left(-y-z\right)}+\frac{b}{y}+\frac{c}{z}=0\)

\(\Leftrightarrow2byz+2cyz+bz^2+cy^2=0\)

Ta lại có:

\(ax^2+by^2+cz^2=\left(-b-c\right)\left(-y-z\right)^2+by^2+cz^2\)

\(=-2byz-2cyz-bz^2-cy^2=0\)

Có:

\(x+y+z=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x=y+z\\-y=x+z\\-z=x+y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=\left(y+z\right)^2\\y^2=\left(x+z\right)^2\\z^2=\left(x+y\right)^2\end{matrix}\right.\)

\(\Rightarrow ax^2+by^2+cz^2\)

\(=a\left(y+z\right)^2+b\left(x+z\right)^2+c\left(x+y\right)^2\)

\(=x^2\left(b+c\right)+y^2\left(a+c\right)+z^2\left(a+b\right)+2\left(ayz+bxz+cxy\right)\)

Mà \(a+b+c=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+c=-a\\a+c=-b\\a+b=-c\end{matrix}\right.\)

Đồng thời có: \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)

\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)

\(\Leftrightarrow ayz+bxz+cxy=0\)

Từ đây ta có:)

\(ax^2+by^2+cz^2=-ax^2-by^2-cz^2\)

\(\Rightarrow2\left(ax^2+by^2+cz^2\right)=0\)

\(\Rightarrow ax^2+by^2+cz^2=0\left(đpcm\right)\)

Do x + y + z = 0 nên

x = - (y + z) ; y = - (x + z) ; z = - (x + y)

=> x² = (y + z)² ; y² = (x + z)² ; z² = (x + y)²

=> ax² + by² + cz² = a(y² + 2yz + z²) + b(x² + 2xz + z²) + c(x² + 2xy + y²) = x²(b + c) + y²(a + c) + z²(a + b) + 2(ayz + bxz + cxy)              (1) 

Thay a = - (b + c) ; b = - (a + c) ; c = - (a + b) (Do a + b + c = 0 ) và ayz+bxz+cxy=0 (do a/x+b/y+c/z=0) vào (1) ta được ax² + by² + cz² = - (ax² + by² + cz²)

=> ax² + by² + cz² = 0

to ko hieu noi