Question

Cho \(\dfrac{a}{b}=\dfrac{c}{d}\left(b+d\ne0\right)\) . Chứng minh: \(\dfrac{4a^2+4c^2}{4b^2+4d^2}=\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\)

Answer 1

Trời tối nên chụp hơi mờ, bạn thông cảm ^^

Answer 2

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow\) a = bk ; c = dk

\(\Rightarrow\)\(\dfrac{4a^2+4c^2}{4b^2+4d^2}\)=\(\dfrac{4\left(bk\right)^2+4\left(dk\right)^2}{4b^2+4d^2}\)

=\(\dfrac{4b^2k^2+4d^2k^2}{4b^2+4d^2}\)=\(\dfrac{k^2\left(4b^2+4d^2\right)}{4b^2+4d^2}\)= k² (1)

\(\Rightarrow\)\(\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\)=\(\dfrac{\left(bk-dk\right)^2}{\left(b-d\right)^2}\)=\(\dfrac{[k\left(b-d\right)]^2}{\left(b-d\right)^2}\)

=\(\dfrac{k^2\left(b-d\right)^2}{\left(b-d\right)^2}\)= k² (2)

Từ (1) và (2), suy ra:

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

Answer 3

Cách 2:

Giải:

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow k^2=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{4a^2+4c^2}{4b^2+4d^2}\) (1)

\(k=\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a-c}{b-d}\Rightarrow k^2=\left(\dfrac{a-c}{b-d}\right)^2=\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\) (2)

Từ (1), (2) \(\Rightarrow\dfrac{4a^2+4c^2}{4b^2+4d^2}=\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\left(đpcm\right)\)

Vậy...