Question

\(M=\frac{2015a}{ab+2015a+2015}+\frac{b}{bc+b+2015}+\frac{c}{ac+c+1}\) biết \(abc=2015\). Tính M.

Answer 1

Ta có:
\(M=\frac{2015a}{ab+2015a+2015}+\frac{b}{bc+b+2015}+\frac{c}{ac+c+1}\)

\(\Rightarrow M=\frac{abca}{ab+abca+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)

\(\Rightarrow M=\frac{abca}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)

\(\Rightarrow M=\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}+\frac{c}{ac+c+1}\)

\(\Rightarrow M=\frac{ac+c+1}{ac+c+1}=1\)

Vậy M = 1

Answer 2

Thay 2015= abc vào M ta được:

M = \(\frac{abca}{ab+abca+abc}\) + \(\frac{b}{bc+b+abc}\) + \(\frac{c}{ac+c+1}\)

M = \(\frac{abca}{ab\left(1+ac+c\right)}\) + \(\frac{b}{b\left(c+1+ac\right)}\) + \(\frac{c}{ac+c+1}\)

M = \(\frac{ac}{1+ac+c}\) + \(\frac{1}{c+1+ac}\) + \(\frac{c}{ac+c+1}\)

M = \(\frac{1+ac+c}{1+ac+c}\) = 1

Vây M = 1

XONG !

Answer 3

Thay abc=2015 vào biểu thức M, ta có:

M=\(\frac{a^2bc}{ab+a^2bc+abc}\)+\(\frac{b}{bc+b+abc}\)+\(\frac{c}{ac+c+1}\)

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

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

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

=1

Vậy M=1

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