Question

Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh rằng \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) ( với giả thiết các tỉ số đều có nghĩa )

Answer 1

Đặt :

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\d=ck\end{matrix}\right.\)

Ta có :

\(VT=\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{d^2}\left(1\right)\)

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

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

Answer 2

Gọi \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=k\(\Rightarrow\)\(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:\(\dfrac{ab}{cd}\)=\(\dfrac{bk.b}{dk.d}\)=\(\dfrac{b^2.k}{d^2.k}\)=\(\dfrac{b^2}{d^2}\)=\(\left(\dfrac{b}{d}\right)^2\)(vì k khác 0) 1

\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)=\(\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}\)=\(\dfrac{\left[b.\left(k+1\right)\right]^2}{\left[d.\left(k+1\right)\right]^2}\)\(\left(\dfrac{b}{d}\right)^2\)=(vì k+1 khác 0) 2

Từ 1 và 2:

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

Vậy \(\dfrac{ab}{cd}\)=\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)(điều cần chứng minh)