Ta có:

\(A=\frac{2015a}{ab+2015a+2015}+\frac{b}{bc+b+2015}+\frac{c}{ac+c+1}\)

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

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

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

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

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

\(\Rightarrow A=1.\)

Vậy \(A=1.\)

Chúc bạn học tốt!

Thay $abc=2015$ vào $A$ ta có:

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

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

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 !

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

CHÚC BẠN HỌC TỐT NHE

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

Ta có

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

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

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

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

ôi câu hỏi hay có khác j câu này Câu hỏi của Lê Phương Thảo - Toán lớp 8 - Học toán với OnlineMath

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

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

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

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

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

\(\Rightarrow M=1\)

Vậy M = 1

abc = 2015 :))

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