a/ Điều kiện xác định \(\hept{\begin{cases}a^2+a\ne0\\a^2-a\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}a\ne0\\a\ne1\\a\ne-1\end{cases}}}\)

b/ \(M=\frac{a^2-1}{2016+2015a^2}\left(\frac{2015a-2016}{a+a^2}+\frac{2016+2015a}{a^2-a}\right)\)

\(=\frac{\left(a-1\right)\left(a+1\right)}{2016+2015a^2}\left(\frac{2015a-2016}{a\left(a+1\right)}+\frac{2016+2015a}{a\left(a-1\right)}\right)\)

\(=\frac{\left(a-1\right)\left(a+1\right)}{2016+2015a^2}\left(\frac{2015a-2016}{a\left(a+1\right)}+\frac{2016+2015a}{a\left(a-1\right)}\right)\)

\(=\frac{\left(a-1\right)\left(a+1\right)}{2016+2015a^2}.\frac{2\left(2015a^2+2016\right)}{a\left(a+1\right)\left(a-1\right)}\)

\(=\frac{2}{a}=\frac{2}{2016}=\frac{1}{1008}\)

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:

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

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

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

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

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

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

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

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

Lời giải:

$\frac{A}{B}=\frac{1}{2}\Rightarrow B=2A$

Khi đó:

$\frac{2015A+1}{2015B+2}=\frac{2015A+1}{2015.2A+2}$

$=\frac{2015A+1}{2(2015A+1)}=\frac{1}{2}=\frac{A}{B}$

Vậy ta có đpcm.

a) Ta có \(a\le b\)

\(\Rightarrow2015a\le2015b\)

\(\Rightarrow2015a-2016\le2015b-2016\)

b) Ta có ​\(a\le b\)​

\(\Rightarrow-a\ge-b\)

\(\Rightarrow-2015a\ge-2015b\)

Xin lỗi mình bấm nhầm

\(\Rightarrow-2015a\ge-2015b\)

\(\Rightarrow-2015a-2017\ge-2015b-2017\)

Mà \(-2015a-2016>-2015a-2017\)

Nên \(-2015a-2016>-2015b-2017\)

ok thanh kiu bạn