Question

cho bốn số a, b, c, d # 0 và thỏa mãn: b²=ac; c²=bd; b³+c³+d³ # 0. CMR: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)

Answer 1

Ta có: \(b^2=ac=>\frac{a}{b}=\frac{b}{c}\)

\(c^2=bd=>\frac{b}{c}=\frac{c}{d}\)

=>\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)

=>\(\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{b}{c}.\frac{b}{c}.\frac{b}{c}=\frac{c}{a}.\frac{c}{a}.\frac{c}{a}=\frac{a}{b}.\frac{b}{c}.\frac{c}{a}\)

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

=>\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a}{d}\)

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

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

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

=>ĐPCM

Answer 2