Question

[Tran Gia Dinh]

CMR: neu a,b,c,d thoa man dang thuc: (ad+bc)^2= 4abcd thi chung lap nen 1 ti le thuc

Answer 1

[KAl(SO4)2·12H2O]

\(\Leftrightarrow ad^2+bc^2=4abcd\Leftrightarrow a^2b^2+b^2c^2+2abcd-4abcd\)

\(\Leftrightarrow a^2b^2-2abcd+b^2d^2=0\)

\(\Leftrightarrow ad^2-bc^2=0\Leftrightarrow ad=bc\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)với \(b;d\ne0\)

Answer 2

[GV]

    \(\left(ad+bc\right)^2=4abcd\)

\(\Leftrightarrow a^2d^2+b^2c^2+2abcd=4abcd\)

\(\Leftrightarrow a^2d^2+b^2c^2+2abcd-4abcd=0\)

\(\Leftrightarrow a^2d^2+b^2c^2-2abcd=0\)

 \(\Leftrightarrow\left(ad-bc\right)^2=0\)

\(\Leftrightarrow ad-bc=0\)

\(\Leftrightarrow ad=bc\)

\(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)  (với điều kiện b, d khác 0)

Vậy a, b, c, d lập thành tỉ lệ thức.