Question

[nữ thám tử nổi tiếng]

cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)

chứng minh

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

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

c.\(\dfrac{2008\cdot a-2009\cdot b}{2009\cdot c+2010\cdot d}=\dfrac{2008\cdot c-2009\cdot d}{2009\cdot a+2010\cdot b}\)

Answer 1

[You Are Mine]

mấy cái đó từ công thức mà ra

Answer 2

[Nguyễn Lê Phước Thịnh]

a: Đặt a/b=c/d=k

=>a=bk; c=dk

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

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

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

b: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)

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

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