Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có :\(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{bk+b}{dk+d}\right)^2=\left(\frac{b\left(k+1\right)}{d\left(k+1\right)}\right)^2=\left(\frac{b}{d}\right)^2\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{bk+b}{dk+d}\right)^2=\left(\frac{b\left(k+1\right)}{d\left(k+1\right)}\right)^2=\left(\frac{b}{d}\right)^2\)
Vậy \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,c=dk\)
Ta có:
\(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{bk+b}{dk+d}\right)^2=\left[\frac{b\left(k+1\right)}{d\left(k+1\right)}\right]^2=\left(\frac{b}{d}\right)^2\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}=\left(\frac{b}{d}\right)^2\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\left(đpcm\right)\)
