Question

Cho \(\dfrac{a}{b}=\dfrac{c}{d}.\) Chứng minh \(\dfrac{a^2-2b^2}{(a+4b)^2}=\dfrac{c^2-2d^2}{(c+4d)^2}\).

Answer 1

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>a=bk; c=dk

\(\dfrac{a^2-2b^2}{\left(a+4b\right)^2}=\dfrac{\left(bk\right)^2-2b^2}{\left(bk+4b\right)^2}=\dfrac{b^2\left(k^2-2\right)}{b^2\left(k+4\right)^2}=\dfrac{k^2-4}{\left(k+4\right)^2}\)

\(\dfrac{c^2-2d^2}{\left(c+4d\right)^2}=\dfrac{\left(dk\right)^2-2d^2}{\left(dk+4d\right)^2}=\dfrac{d^2\left(k^2-2\right)}{d^2\left(k+4\right)^2}=\dfrac{k^2-2}{\left(k+4\right)^2}\)

Do đó: \(\dfrac{a^2-2b^2}{\left(a+4b\right)^2}=\dfrac{c^2-2d^2}{\left(c+4d\right)^2}\)