Question

Cho\(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{d}\). Chứng minh rằng( \(\dfrac{a+b+c}{b+c+d}\))^3=\(\dfrac{a}{d}\)
Cảm ơn!

Answer 1

Ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(1\right)\)

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:

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

Answer 2

Ta có \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(1\right)\)

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

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

Từ \(\left(1\right),\left(2\right)\) =>\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)(đpcm)

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