Chủ đề này có 1 trả lời

[tpdtthltvp]

Cho $\frac{x^2-yz}{a}=\frac{y^2-xz}{b}=\frac{z^2-xy}{c}$. Chứng minh rằng $\frac{a^2-bc}{x}=\frac{b^2-ac}{y}=\frac{c^2-ab}{z}$

[chaubee2001]

Giải:

Ta có:

$\Leftrightarrow \frac{a}{x^2-yz}=\frac{b}{y^2-xz}=\frac{c}{z^2-xy}$

$\Leftrightarrow \frac{a^2}{(x^2-yz)^2}=\frac{bc}{(y^2-xz)(z^2-xy)} \Leftrightarrow \frac{a^2}{(x^2-yz)^2}=\frac{bc}{(y^2-xz)(z^2-xy)}=\frac{a^2-bc}{x^4-2x^2yz+y^2z^2-(y^2z^2-xy^3-xz^3+x^2yz)}=\frac{a^2-bc}{x^4+xy^3+xz^3-3x^2yz}=\frac{a^2-bc}{x(x^3+y^3+z^3-3xyz)}$

Tương tự, $\frac{b^2}{(y^2-xz)^2}=\frac{b^2-ac}{y(x^3+y^3+z^3-3xyz)}$,$\frac{c2}{(z^2-xy)^2}=\frac{b^2-ac}{z(x^3+y^3+z^3-3xyz)}$

Nên $\frac{a^2-bc}{x}=\frac{b^2-ac}{y}=\frac{c^2-ab}{z}$(đpcm)