Question

Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) CMR:

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

  c) \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)

Đừng dùng hằng đẳng thức nha mn

Answer 1

a: Đặt a/b=c/d=k

=>a=bk; c=dk

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

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

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

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

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

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

 

Answer 2

a) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow a=bk,c=dk\)

\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)

\(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)

\(\Rightarrowđpcm\)