Cho ${{x}_{1}},\,{{x}_{2}},\,{{x}_{3}},\,{{x}_{4}}\ne 0$ và ${{x}_{2}}^{2}={{x}_{1}}.{{x}_{3}}$; ${{x}_{3}}^{2}={{x}_{2}}.{{x}_{4}}$

Chứng minh rằng $\frac{{{x}_{1}}^{3}+{{x}_{2}}^{3}+{{x}_{3}}^{3}}{{{x}_{2}}^{3}+{{x}_{3}}^{3}+{{x}_{4}}^{3}}=\frac{{{x}_{1}}}{{{x}_{4}}}$
Giải

Ta có : ${{x}_{2}}^{2}={{x}_{1}}.{{x}_{3}}\Leftrightarrow \frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{x}_{2}}}{{{x}_{3}}}$

${{x}_{3}}^{2}={{x}_{2}}.{{x}_{4}}\Leftrightarrow \frac{{{x}_{3}}}{{{x}_{4}}}=\frac{{{x}_{2}}}{{{x}_{3}}}$

$\Rightarrow \frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{x}_{2}}}{{{x}_{3}}}=\frac{{{x}_{3}}}{{{x}_{4}}}\Rightarrow {{\left( \frac{{{x}_{1}}}{{{x}_{2}}} \right)}^{3}}={{\left( \frac{{{x}_{2}}}{{{x}_{3}}} \right)}^{3}}={{\left( \frac{{{x}_{3}}}{{{x}_{4}}} \right)}^{3}}\Leftrightarrow \frac{{{x}_{1}}^{3}}{{{x}_{2}}^{3}}=\frac{{{x}_{2}}^{3}}{{{x}_{3}}^{3}}=\frac{{{x}_{3}}^{3}}{{{x}_{4}}^{3}}$

Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{{{x}_{1}}^{3}}{{{x}_{2}}^{3}}=\frac{{{x}_{2}}^{3}}{{{x}_{3}}^{3}}=\frac{{{x}_{3}}^{3}}{{{x}_{4}}^{3}}=\frac{{{x}_{1}}^{3}+{{x}_{2}}^{3}+{{x}_{3}}^{3}}{{{x}_{2}}^{3}+{{x}_{3}}^{3}+{{x}_{4}}^{3}}$ (1)

Mà $\frac{{{x}_{1}}^{3}}{{{x}_{2}}^{3}}=\frac{{{x}_{1}}}{{{x}_{2}}}.\frac{{{x}_{1}}}{{{x}_{2}}}.\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{x}_{1}}}{{{x}_{2}}}.\frac{{{x}_{2}}}{{{x}_{3}}}.\frac{{{x}_{3}}}{{{x}_{4}}}=\frac{{{x}_{1}}}{{{x}_{4}}}$ (2)

Từ (1) và (2) suy ra $\frac{{{x}_{1}}^{3}+{{x}_{2}}^{3}+{{x}_{3}}^{3}}{{{x}_{2}}^{3}+{{x}_{3}}^{3}+{{x}_{4}}^{3}}=\frac{{{x}_{1}}}{{{x}_{4}}}$