Question

Cho ac=bd. Chứng minh:\(\left(\frac{a+d}{b+c}\right)^2=\frac{a^2+d^2}{b^2+c^2}\)

Answer 1

a/b =c/d =k,a=bk,c=dk;bkdk/bd =k²(1)

(bk+dk)²   /(b+d)² =(k(b+d))²/(b+d)² =k²(b+d)² / (b+d)² = k²(2)

từ (1) và (2) suy ra điều cần chứng minh.

Answer 2

\(\text{Ta có:}ac=bd\Rightarrow\frac{a}{b}=\frac{d}{c}=\frac{a+d}{b+c}\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{d}{c}\right)^2=\left(\frac{a+d}{b+c}\right)^2\)

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

\(\text{Mặt khác: }\frac{a^2}{b^2}=\frac{d^2}{c^2}=\frac{a^2+d^2}{b^2+c^2}\)

\(\text{Suy ra: }\left(\frac{a+d}{b+c}\right)^2=\frac{a^2+d^2}{b^2+c^2}\)