Question

Cho biết ax + by + cz = 0

Rút gọn: \(A=\frac{bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2}{ax^2+by^2+cz^2}\)

Answer 1

Giải

Ta có: \(B=bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2\)

\(=bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2\left(bcyz+acxz+abxy\right)\)

\(=ax^2\left(b+c\right)+by^2\left(a+c\right)+cz^2\left(a+b\right)-2\left(bcyz+acxz+abxy\right)\)(1)

Từ giả thiết suy ra:

\(a^2x^2+b^2y^2+c^2z^2+2\left(abxy+acxz+bcyz\right)=0\) (2)

Từ (1) và (2):

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

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

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

Do đó:

\(A=\frac{B}{ax^2+by^2+cz^2}=a+b+c\)