Question

Bài 1:Cho \(\dfrac{a}{k}=\dfrac{x}{a};\dfrac{b}{k}=\dfrac{y}{b}\).CMR: \(\dfrac{a^2}{b^2}=\dfrac{x}{y}\)

Bài 2: Cho a=b+c và c=\(\dfrac{bd}{b-d}\) \(\left(b\ne0;d\ne0\right)\)

CMR:\(\dfrac{a}{b}=\dfrac{c}{d}\)

Answer 1

BÀI 1:

\(\dfrac{a}{k}=\dfrac{x}{a}\Rightarrow a^2=kx\)

\(\dfrac{b}{k}=\dfrac{y}{b}\Rightarrow b^2\)=ky

Vay \(\dfrac{a^2}{b^2}=\dfrac{kx}{ky}=\dfrac{x}{y}\)

Answer 2

Bài 2:

Vì a=b+c nên ad=(b+c)d=bd+cd (1)

Vi c=\(\dfrac{bd}{b-d}\)nen \(bd=\)c.(b-d)=bc-cd hay bc=bd+cd (2)

Từ (1),(2) =>ad=bc=>\(\dfrac{a}{b}=\dfrac{c}{d}\)