Question

C/minh nếu \(\frac{a}{b}=\frac{c}{d}\)thì \(\frac{a^2+b^2}{c^2+d^2}=\frac{a.b}{c.d}\)

 

Answer 1

Đặt \(\frac{a}{b}=\frac{c}{d}=k=>a=bk,c=dk\)

Ta có:\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}\)

\(\frac{a.b}{c.d}=\frac{bk.b}{dk.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\)

=>\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2}{d^2}=\frac{a.b}{c.d}\)

=>\(\frac{a^2+b^2}{c^2+d^2}=\frac{a.b}{c.d}\)