so sánh A=1 phần 1*2*3 + 1 phần 2*3*4 +...+1 phần 2015*2016*2017 với 1 phần 4
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DK
1
NN
4 tháng 5 2018
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2015.2016.2017}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2015.2016.2017}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...-\frac{1}{2016.2017}\)
\(=\frac{1}{2}-\frac{1}{2016.2017}< \frac{1}{2}\)
\(\Rightarrow2A< \frac{1}{2}\Rightarrow A< \frac{1}{4}\)
19 tháng 8 2018
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
KM
4
HP
13 tháng 5 2016
\(A=\frac{10^{2015}+1}{10^{2016}+1}\Rightarrow10A=\frac{10.\left(10^{2015}+1\right)}{10^{2016}+1}=\frac{10^{2016}+10}{10^{2016}+1}\)
\(A=\frac{10^{2016}+1+9}{10^{2016}+1}=\frac{10^{2016}+1}{10^{2016}+1}+\frac{9}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)
\(B=\frac{10^{2016}+1}{10^{2017}+1}\Rightarrow10B=\frac{10.\left(10^{2016}+1\right)}{10^{2017}+1}=\frac{10^{2017}+10}{10^{2017}+1}\)
\(B=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)
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
13 tháng 5 2016
\(B=\frac{10^{2016}+1}{10^{2017}+1}<\frac{10^{2016}+1+9}{10^{2017}+1+9}\)
\(=\frac{10^{2016}+10}{10^{2017}+10}\)
\(=\frac{10.\left(10^{2015}+1\right)}{10.\left(10^{2016}+1\right)}\)
\(=\frac{10^{2015}+1}{10^{2016}+1}=A\)
\(\Rightarrow\) B<A
TN
1
PT
2
17 tháng 1 2017
\(\frac{2015}{2016}+\frac{2016}{2017}>\frac{\left(2015+2016\right)}{\left(2016+2017\right)}=\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)
BD
0
10 tháng 7 2017
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