Question

​Cho a/b=c/d và b+d khac 0.CMR: 3a²+c²/3b²+d²=(a+c)²/(b+d)²

Answer 1

Đặt\(\frac{a}{b}=\frac{c}{d}=k\left(k\in Q\right)\)

\(\Rightarrow\hept{\begin{cases}a=bk\left(1\right)\\c=dk\left(2\right)\end{cases}}\)

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

Thay \(\left(1\right),\left(2\right)vào\left(3\right)có\)

\(\frac{3b^2k^2+d^2k^2}{3b^2+d^2}=\frac{k^2\left(3b^2+d^2\right)}{3b^2+d^2}=k^2\left(4\right)\)

\(\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\frac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\left(5\right)\)

Từ \(\left(4\right),\left(5\right)\Rightarrowđpcm\)