Ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{aa}{bb}=\dfrac{a^2+a^2}{b^2+b^2}\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{a^2.2}{b^2.2}\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{a^2}{b^2}\)
\(\Leftrightarrow\dfrac{ac}{bd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\rightarrowđpcm\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\)
VT: \(\dfrac{ac}{bd}=\dfrac{kb.kd}{b.d}=k^2\) (1)
VP: \(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(kb+kd\right)^2}{\left(b+d\right)^2}=\dfrac{\left[k.\left(b+d\right)\right]^2}{\left(b+d\right)^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\) (2)
Từ (1) và (2), suy ra:
\(\dfrac{ac}{bd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\) (đpcm)
Theo bài ra ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^2=\left(\dfrac{a+c}{b+d}\right)^2\) \(\left(1\right)\)
Theo bài ra ta lại có : \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^2=\dfrac{ac}{bd}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra : \(\left(\dfrac{a+c}{b+d}\right)^2=\left(\dfrac{a}{b}\right)^2=\dfrac{ab}{cd}\)
\(\Rightarrow\left(\dfrac{a+c}{b+d}\right)^2=\dfrac{ab}{cd}\left(ĐPCM\right)\)
Vậy \(\left(\dfrac{a+c}{b+d}\right)^2=\dfrac{ab}{cd}\)
