Question

Cho \(\dfrac{ab}{2014}=\dfrac{1}{c}\)

Tính giá trị của biểu thức

\(A=\dfrac{2014a}{ab+2014a+2014}+\dfrac{b}{bc+b+2014}+\dfrac{c}{ac+c+1}\)

Answer 1

Từ \(\dfrac{ab}{2014}=\dfrac{1}{c}\Rightarrow abc=2014\) thay vào \(A\) ta có:

\(A=\dfrac{abc\cdot a}{ab+abc\cdot a+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{a^2bc}{ab+a^2bc+abc}+\dfrac{b}{b\left(ac+c+1\right)}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{ac\cdot ab}{ab\left(ac+c+1\right)}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{ac+c+1}{ac+c+1}=1\Rightarrow A=1\)