Question

 \(cho\frac{a}{b}=\frac{b}{c}=\frac{c}{d}.CMR:\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)

Answer 1

Có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)(Tính chất dãy tỉ số bằng nhau)

=> \(\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{c^2}{d^2}=\frac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}=\frac{abc}{bcd}=\frac{a}{d}\)

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

=> Đpcm

Answer 2

Ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\frac{abc}{bcd}=\frac{a}{d}\)( theo TC dãy TSBN)

Mà:\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\)(theo TC dãy TSBN)

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