Question

Cho a, b, c \(\in\) R và a, b, c \(\ne\) 0 thỏa mãn b² = ac. Chứng minh rằng :

\(\dfrac{a}{c}=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}\) (Biết rằng các tỉ số đều có nghĩa)

Answer 1

Giải:

Ta có: \(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=k\)

+) \(k^2=\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a}{c}\) (1)

+) \(k=\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{2011b}{2011c}=\dfrac{a+2011b}{b+2011c}\) ( t/c dãy tỉ số bằng nhau )

\(\Rightarrow k^2=\left(\dfrac{a+2011b}{b+2011c}\right)^2=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}\) (2)

Từ (1), (2) \(\Rightarrow\dfrac{a}{c}=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}\left(đpcm\right)\)

Answer 2

Giải:

Từ hằng đẳng thức: \(\left(a+b\right)^2=a^2+2ab+b\) ta có:

\(VP=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}=\dfrac{a^2+2.2011ab+\left(2011b\right)^2}{b^2+2.2011bc+\left(2011c\right)^2}\)

\(=\dfrac{a^2+2.2011ab+2011^2ac}{ac+2.2011bc+2011^2c^2}\)

\(=\dfrac{a\left(a+2.2011b+2011^2c\right)}{c\left(a+2.2011b+2011^2c\right)}=\dfrac{a}{c}=VT\)

Vậy \(\dfrac{a}{c}=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}\) (Đpcm)