Question

cho tỉ lệ thức \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\) chứng minh:

a. \(\dfrac{a}{b}\)=\(\dfrac{a+c}{b+d}\) ( với b+d ≠ 0)

b. \(\dfrac{ac}{bd}\)=\(\dfrac{a^2+c^2}{b^2+d^2}\)

Answer 1

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}\)=k (1)

=> a=bk ,c=dk

a.Có \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\left(2\right)\)

Từ (1) và (2)=>\(\dfrac{a+c}{b+d}=\dfrac{a}{b}\left(=k\right)\)

b. Có \(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\)

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

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