Question

Bài 1 Cho \(a^2\)=bc. Chứng minh rằng

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

Answer 1

Giải:

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

a, Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

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

\(\Rightarrowđpcm\)

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

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

\(\Rightarrowđpcm\)