Question

Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh các tỉ lệ thức sau:

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

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

Answer 1

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{a^2-b^2}{ab}=\dfrac{b^2k^2-b^2}{bk\cdot b}=\dfrac{b^2\left(k^2-1\right)}{b^2k}=\dfrac{k^2-1}{k}\)

\(\dfrac{c^2-d^2}{cd}=\dfrac{d^2k^2-d^2}{dk\cdot d}=\dfrac{d^2\left(k^2-1\right)}{d^2\cdot k}=\dfrac{k^2-1}{k}\)

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

b: \(\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(bk+b\right)^2}{b^2k^2+b^2}=\dfrac{b^2\cdot\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\dfrac{\left(k+1\right)^2}{k^2+1}\)

\(\dfrac{\left(c+d\right)^2}{c^2+d^2}=\dfrac{\left(dk+d\right)^2}{d^2k^2+d^2}=\dfrac{\left(k+1\right)^2}{k^2+1}\)

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