Question

Cho a, b, c, d, e khác 0 thỏa mãn điều kiện \(b^2=ac;c^2=bd;d^2=ce\). Chứng minh rằng : \(\dfrac{a^4+b^4+c^4+d^4}{b^4+c^4+d^4+e^4}\)=\(\dfrac{a}{e}\)

Answer 1

Lời giải:

Từ \(b^2=ac; c^2=bd; d^2=ce\)

\(\Rightarrow \frac{b}{a}=\frac{c}{b}; \frac{c}{b}=\frac{d}{c}; \frac{d}{c}=\frac{e}{d}\)

\(\Rightarrow \frac{b}{a}=\frac{c}{b}=\frac{d}{c}=\frac{e}{d}\).

Đặt \( \frac{b}{a}=\frac{c}{b}=\frac{d}{c}=\frac{e}{d}=k\Rightarrow b=ak; c=bk; d=ck; e=dk\)

Khi đó:

\(\frac{a^4+b^4+c^4+d^4}{b^4+c^4+d^4+e^4}=\frac{a^4+b^4+c^4+d^4}{a^4k^4+b^4k^4+c^4k^4+d^4k^4}=\frac{a^4+b^4+c^4+d^4}{k^4(a^4+b^4+c^4+d^4)}=\frac{1}{k^4}(1)\)

Và: \(bcde=ak.bk.ck.dk\)

\(\Rightarrow e=ak^4\Rightarrow \frac{a}{e}=\frac{1}{k^4}(2)\)

Từ \((1);(2)\Rightarrow \frac{a^4+b^4+c^4+d^4}{b^4+c^4+d^4+e^4}=\frac{a}{e}\)