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

Vì \(\dfrac{a}{c}=\dfrac{c}{b}\Rightarrow\left(\dfrac{a}{c}\right)^2=\left(\dfrac{c}{b}\right)^2\Rightarrow\dfrac{a^2}{c^2}=\dfrac{c^2}{b^2}\)

Theo t/chất dãy tỉ số = nhau, ta có:

\(\dfrac{a^2}{c^2}=\dfrac{c^2}{b^2}=\dfrac{a^2+c^2}{c^2+b^2}\)\(\Rightarrow a\left(c^2+b^2\right)=b\left(a^2+c^2\right)\Rightarrow ac^2+ab^2=ba^2+bc^2\)

\(\Rightarrow ab^2=ba^2+bc^2-ac^2\)

\(\Rightarrow ab^2-a^3=\left(ba^2+bc^2\right)-\left(a^3+ac^2\right)\)

\(\Rightarrow a\left(b^2-a^2\right)=\left(b-a\right)\left(a^2+c^2\right)\)

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

Vậy...

Bài này đúng á, mk học rồi, chúc p hk tốt