Question

Cho biết a+c=2b;và 2bd=c(b+d) , chứng minh rằng: \(2\left(\frac{10a+c}{10b+d}\right)^2-\left(\frac{a}{b}\right)^2=\left(\frac{c}{d}\right)^2\)

Answer 1

\(a+c=2b\Rightarrow2bd=ad+cd=c\left(b+d\right)=bc+cd\)

\(\Rightarrow ad=bc\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Lúc đó: \(2\left(\frac{10a+c}{10b+d}\right)^2-\left(\frac{a}{b}\right)^2=2\left(\frac{10.bk+dk}{10b+d}\right)^2-\left(\frac{bk}{b}\right)^2\)

\(=2k^2-k^2=k^2\)(1)

và \(\left(\frac{c}{d}\right)^2=\left(\frac{dk}{d}\right)^2=k^2\)(2)

Từ (1) và (2) suy ra \(2\left(\frac{10a+c}{10b+d}\right)^2-\left(\frac{a}{b}\right)^2=\left(\frac{c}{d}\right)^2\)(đpcm)