Question

Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\), chứng minh rằng

\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

 

Answer 1

ta có:\(\frac{a}{b}\)=\(\frac{c}{d}\)=k

\(\Rightarrow\)a=bk;c=dk

ta có:\(\frac{a.b}{cd}\)=\(\frac{bk.b}{dk.d}\)=\(\frac{kb^2}{kd^2}\)=\(\frac{b^2}{d^2}\)

ta có:\(\frac{a^2+b^2}{c^2+d^2}\)=\(\frac{k^2.b^2+b^2}{k^2.d^2+d^2}\)=\(\frac{b^2(k+1)}{d^2(k+1)}\)=\(\frac{b^2}{d^2}\)

vậy:\(\frac{a^2+b^2}{c^2+d^2}\)\(=\)\(\frac{ab}{cd}\)