Question

cho TLT \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh

\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)

\(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)

Answer 1

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\) 

_________________

Ta có: 

\(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{dk+c}{dk-c}=\dfrac{c+d}{c-d}\) (đpcm) 

_________________

Ta có:

\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{a^2+b^2}{c^2+d^2}\) (đpcm)