Question

[Phạm Minh Hiếu]

CHO \(\frac{a}{b}=\frac{c}{d}\)và \(b+d\ne0\).CHỨNG TỎ : \(\frac{3a^2+c^2}{3b^2+d^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

Answer 1

[Phạm Tuấn Đạt]

Gọi \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=kb;c=kd\)(1)

Thay (1) vào ta có :

\(\frac{3a^2+c^2}{3b^2+d^2}=\frac{3\left(kb\right)^2+\left(kd\right)^2}{3b^2+d^2}=\frac{3k^2b^2+k^2+d^2}{3b^2+d^2}=\frac{k^2\left(3b^2+d^2\right)}{3b^2+d^2}=k^2\)(1)

\(\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(kb+kd\right)^2}{\left(b+d\right)^2}=\frac{\left[k\left(b+d\right)\right]^2}{\left(b+d\right)^2}=\frac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\)(2)

Từ (1) và (2)

\(\Rightarrow\frac{3a^2+c^2}{3b^2+d^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

\(\RightarrowĐPCM\)