Đặt: \(\frac{a}{b}=\frac{c}{d}=k\)
==> a = b.k
c = d.k
Ta có : \(\frac{a^2+b^2}{c^2+d^2}\) = \(\frac{b^2.k^2+b^2}{d^2.k^2+d^2}\) = \(\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}\) = \(\frac{b^2}{d^2}\) (1)
\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) = \(\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\) = \(\frac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}\) = \(\frac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}\) = \(\frac{b^2}{d^2}\) (2)
Từ (1) và (2) ==> \(\frac{a^2+b^2}{c^2+d^2}\) = \(\frac{\left(a-b\right)^2}{\left(c-d^{ }\right)^2}\) (đpcm)
Good for you