Question

cho \(\dfrac{a}{b}\)chứng minh rằng :

\(a,\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\\ b,\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)

Answer 1

a,Từ \(\dfrac{a}{c}=\dfrac{c}{b}\)⇒\(c^2=a.b\)

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

Answer 2

b,Ta có:

\(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\Rightarrow\dfrac{b^2+c^2}{a^2+c^2}=\dfrac{a}{b}\\ \dfrac{a^2+c^2}{b^2+c^2}=\dfrac{b}{a}\Rightarrow\dfrac{b^2+c^2}{a^2+c^2}-1=\dfrac{b}{a}-1\\ hay\dfrac{b^2+c^2-a^2-c^2}{a^2+c^2}=\dfrac{b-a}{a}\)

Vậy \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)