Question

\(\frac{a}{b}=\frac{c}{d}\)(a,b,c,d khác 0)

Chứng minh :\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

Answer 1

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