Question

Từ tỉ lệ thức\(\dfrac{a-b}{a+b}\)=\(\dfrac{b-c}{b+c}\), chứng minh b² = a.c

Answer 1

\(\dfrac{a-b}{a+b}=\dfrac{b-c}{b+c}\\ \Rightarrow\left(a-b\right)\left(b+c\right)=\left(b-c\right)\left(a+b\right)\\ \Leftrightarrow a\left(b+c\right)-b\left(b+c\right)=b\left(a+b\right)-c\left(a+b\right)\\ \Leftrightarrow ab+ac-b^2-bc=ab+b^2-ca-cb\\ \Leftrightarrow ab-ab+ac+ac=b^2+b^2+bc-bc\\ \Leftrightarrow2ac=2b^2\\ \Leftrightarrow ac=b^2\)

Answer 2

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

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

Áp dụng t/c của dãy tỉ số bằng nhau ta cố:

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

=>\(b^2=ac\left(dpcm\right)\)