Question

cho \(\frac{a}{b}\)=\(\frac{c}{d}\)

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

Answer 1

Đăt \(\frac{a}{b}=\frac{c}{d}=k\)\(\Rightarrow a=bk;c=dk\)

Khi đó \(\frac{3a^2+c^2}{3b^2+d^2}=\frac{3.\left(bk\right)^2+\left(dk^2\right)}{3.b^2+d^2}=\frac{3b^2.k^2+d^2.k^2}{3b^2+d^2}=\frac{k^2.\left(3b^2+d^2\right)}{3b^2+d^2}=k^2\) (1)

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

Từ (1) và (2) ta có \(\frac{3a^2+c^2}{3b^2+d^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

Answer 2

ra bằng 

Answer 3

Áp dụng dãy tỉ số bằng nhau ta có:

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

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