Question

Cho\(\frac{a}{b}\)=\(\frac{c}{d}\),chứng minh rằng \(\frac{ab}{cd}\)=\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\) 

 

Answer 1

Cho \(\frac{a}{b}\)=\(\frac{c}{d}\)=k=> a=bk; c=dk

vế trái =\(\frac{ab}{cd}\)=\(\frac{bk.b}{dk.d}\)=\(\frac{b^2}{d^2}\)(1)

vế phải =\(\frac{\left(a+b\right)2}{\left(c+d\right)^2}\)=\(\frac{\left(bk+b\right)2}{\left(dk+d\right)^2}\)=\(\frac{b^2\left(k+1\right)2}{d^2\left(k+1\right)^2}\)=\(\frac{b^2}{d^2}\)(2)

Từ (1) và (2) ta có:\(\frac{ab}{cd}\)=\(\frac{\left(a+b\right)2}{\left(c+d\right)^2}\)

Answer 2

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)