Question

[Thu Huyền]

Cho \(ax+by+cz=0\)

Rút gọn \(\dfrac{bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2}{ax^2+by^2+cz^2}\)

Answer 1

[Mashiro Shiina]

\(\dfrac{bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2}{ax^2+by^2+cz^2}\)

\(=\dfrac{bcy^2+bcz^2+acx^2+acz^2+abx^2+aby^2-2\left(bcyz+acxz+abxy\right)}{ax^2+by^2+cz^2}\)

\(=\dfrac{bcy^2+bcz^2+acx^2+acz^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2-a^2x^2-b^2y^2-c^2z^2-2\left(bcyz+acxz+abxy\right)}{ax^2+by^2+cz^2}\)

\(=\dfrac{ax^2\left(a+b+c\right)+by^2\left(a+b+c\right)+cz^2\left(a+b+c\right)-\left(ax+by+cz\right)^2}{ax^2+by^2+cz^2}\)

\(=\dfrac{\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)}{ax^2+by^2+cz^2}=a+b+c\)