Đã gửi bởi
Cho $\frac{ay-bx}{c}=\frac{cx-az}{b}=\frac{bz-cy+bc}{a}$. Chứng minh rằng $(a^{2}+b^{2}+c^{2})^{2}(ay-bx)(cx-az)=a^{2}b^{3}c^{3}$
$\frac{ay-bx}{c}=\frac{cx-az}{b}=\frac{bz-cy+bc}{a} =\frac{cay-cbx}{c^2}=\tfrac{bcx-baz}{b^2}=\frac{abz-acy+abc}{a^2}=\frac{-abc}{a^2+b^2+c^2}\Rightarrow \frac{a^2b^2c^2}{(a^2+b^2+c^2)^2}=\frac{(ay-bx)(cx-az)}{bc}\Rightarrow (a^2+b^2+c^2)^2(ay-bx)(cx-az)=a^2b^3c^3$
