Question

1. Cho a, b, c, d là 4 số khác 0 thoả mãn điều kiện b² = a.c ; c² = b.d và b³ + c³ d³ khác 0

Chứng minh rằng: a³ + b³ + c³ : b³ + c³ + d³ = \(\dfrac{a}{b}\)

HELP ME

Answer 1

Ta có:

\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)

\(c^2=b.d\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\)

Do đó:\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

Do đó:\(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{b}\left(đpcm\right)\)

Answer 2

\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

Vậy \(\dfrac{a}{b}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

\(\rightarrowđpcm\)