\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2015.2016.2017}\)

\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2015.2016.2017}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...-\frac{1}{2016.2017}\)

\(=\frac{1}{2}-\frac{1}{2016.2017}< \frac{1}{2}\)

\(\Rightarrow2A< \frac{1}{2}\Rightarrow A< \frac{1}{4}\)

A,(2016*2017-1)/2015*2017+2016

   =2015*2017+2016/2015*2017+2016

   =1

B,18*(1919/2121+888/999)

  =18*(19/21+8/9)

  =18*(19/21)+18*(8/9)=114/7+16=16+2/7+16=32/2/7

c,gọi s=1/2+1/4+...+1/32

2s=1+1/2+...+1/16

2s-s=(1+1/2+...+1/16)-(1/2+1/4+...+1/32)

s=1-1/32=31/32

\(A=\frac{10^{2015}+1}{10^{2016}+1}\Rightarrow10A=\frac{10.\left(10^{2015}+1\right)}{10^{2016}+1}=\frac{10^{2016}+10}{10^{2016}+1}\)

\(A=\frac{10^{2016}+1+9}{10^{2016}+1}=\frac{10^{2016}+1}{10^{2016}+1}+\frac{9}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)

\(B=\frac{10^{2016}+1}{10^{2017}+1}\Rightarrow10B=\frac{10.\left(10^{2016}+1\right)}{10^{2017}+1}=\frac{10^{2017}+10}{10^{2017}+1}\)

\(B=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)

Vì 10²⁰¹⁶+1 < 10²⁰¹⁷+1

=>\(\frac{9}{10^{2016}+1}>\frac{9}{10^{2017}+1}\)

=>\(1+\frac{9}{10^{2016}+1}>1+\frac{9}{10^{2017}+1}\)

=>10A > 10B

=>A > B

\(B=\frac{10^{2016}+1}{10^{2017}+1}<\frac{10^{2016}+1+9}{10^{2017}+1+9}\)

      \(=\frac{10^{2016}+10}{10^{2017}+10}\)

      \(=\frac{10.\left(10^{2015}+1\right)}{10.\left(10^{2016}+1\right)}\)

      \(=\frac{10^{2015}+1}{10^{2016}+1}=A\)

\(\Rightarrow\) B<A

\(10A=\frac{10\left(10^{2015}+1\right)}{10^{2016}+1}=\frac{10^{2016}+10}{10^{2016}+1}=\frac{10^{2016}+1+9}{10^{2016}+1}=\frac{10^{2016}+1}{10^{2016}+1}+\frac{9}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)

\(10B=\frac{10\left(10^{2016}+1\right)}{10^{2017}+1}=\frac{10^{2017}+10}{10^{2017}+1}=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)

vì 10²⁰¹⁶+1<10²⁰¹⁷+1

=>\(\frac{9}{10^{2016}+1}>\frac{9}{10^{2017}+1}\)

=>A>B

a kiếm phân số trung gian để so sánh

\(\frac{2015}{2016}+\frac{2016}{2017}>\frac{\left(2015+2016\right)}{\left(2016+2017\right)}=\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)

ko bit

a) \(\frac{x+2015}{5}+\frac{x+2015}{6}=\frac{x+2015}{7}+\frac{x+2015}{8}\)

\(\frac{x+2015}{5}+\frac{x+2015}{6}-\frac{x+2015}{7}-\frac{x+2015}{8}=0\)

\(\left(x+2015\right).\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\)

vì \(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\ne0\)

\(\Rightarrow\)x + 2015 = 0

\(\Rightarrow\)x = -2015

b) Tương tự