Question

Cho \(\dfrac{a}{c}=\dfrac{c}{b}\) chứng minh rằng:

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

Answer 1

Từ \(\dfrac{a}{c}=\dfrac{c}{b}\Rightarrow c^2=ab\)

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

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

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

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

Answer 2

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