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P \(=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
P\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{50^2-1}{50^2}\)
P \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
P\(=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
P\(=\frac{1.51}{50.2}=\frac{51}{100}\)
1. \(G=2016.2016=\left(2014+2\right)\left(2018-2\right)=2014.2018-4028+4036-4=2014.2018+4\)
vì 2014.2018+4 >2014.2018
=> G>H
\(\frac{2016.2016}{2013.2019}=\frac{\left(2013+3\right)\left(2019-3\right)}{2013.2019}=\frac{2013.2019-6039+6057-9}{2013.2019}=\frac{2013.2019+9}{2013.2019}=1+\frac{9}{2013.2019}\)
vì \(1+\frac{9}{2013.2019}>1\)
\(\frac{2016.2016}{2013.2019}>1\)
\(2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{2016}{2^{2015}}\)
\(2A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}-\frac{2016}{2^{2016}}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}-\frac{1}{2^{2016}}< 1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)(1)
Ta có
\(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^{2014}}-\frac{1}{2^{2015}}\right)=1+\left(1-\frac{1}{2^{2015}}\right)\)
\(< 1+1=2\)(2)
Từ (1) và (2) ta có A<2
Vậy A<B
A=\(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+.........+\frac{2016}{2^{2016}}\\ 2A=1+\frac{2}{2}+\frac{3}{2^2}+........+\frac{2016}{2^{2015}}\\ 2A-A=\left(\frac{2}{2}-\frac{1}{2}\right)+\left(\frac{3}{2^2}-\frac{2}{2^2}\right)+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+.........\left(\frac{2016}{2^{2015}}-\frac{2015}{2^{2015}}\right)+\left(1-\frac{2016}{2^{2015}}\right)\\ A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2015}}+\left(1-\frac{2016}{2^{2015}}\right)\)
\(GọiC=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2015}}\\ 2C=1+\frac{1}{2}+\frac{1}{2^3}+......+\frac{1}{2^{2014}}\\ 2C-C=C=1-\frac{1}{2^{2015}}\)
Thay C vào A , ta có : A = 1 - 1/2^2015 + 1 - 1/2^2016 =2 - 1/2^2015 - 1/2^2016<2 =B->A<B
\(A=\frac{2016^{2016}-1+3}{2016^{2016}-1};B=\frac{2016^{2016}-3+3}{2016^{2016}-3}\)
\(A=\frac{2016^{2016}-1}{2016^{2016}-1}+\frac{3}{2016^{2016}-1};B=\frac{2016^{2016}-3}{2016^{2016}-3}+\frac{3}{2016^{2016}-3}\)
\(A=1+\frac{3}{2016^{2016}-1};B=1+\frac{3}{2016^{2016}-3}\)
Vì \(\frac{3}{2016^{2016}-1}< \frac{3}{2016^{2016}-3}\)
\(\Rightarrow1+\frac{3}{2016^{2016}-1}< 1+\frac{3}{2016^{2016}-3}\)
\(\Rightarrow A< B\)
\(A=\frac{2016^{2016}+2}{2016^{2016}-1}=\frac{2016^{2016}-1+3}{2016^{2016}-1}=1+\frac{3}{2016^{2016}-1}\)
\(B=\frac{2016^{2016}}{2016^{2016}-3}=\frac{2016^{2016}-3+3}{2016^{2016}-3}=1+\frac{3}{2016^{2016}-3}\)
Do \(\frac{3}{2016^{2016}-1}>\frac{3}{2016^{2016}-3}\)
\(\Rightarrow1+\frac{3}{2016^{2016}-1}>1+\frac{3}{2016^{2016}-3}\)
\(\Rightarrow A>B\)
Vậy \(A>B\)
Chúc bạn học tốt !!!
Bài 1:
ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{100^2}\)
\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
Bài 2:
ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Học tốt nhé bn !!
\(A=2^0+2^1+2^3+.....+2^{2016}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{2017}\)
\(\Rightarrow A=2A-A=2^{2017}-1\)
\(\Rightarrow A=B\)
Mình ko bít có đúng ko nên sai đừng trách mình nhé !
\(A=\frac{7^{2011}+1}{7^{2013}+1}\)
\(7^2.A=\frac{7^{2013}+49}{7^{2013}+1}=\frac{7^{2013}+1+48}{7^{2013}+1}=\)\(\frac{7^{2013}+1}{7^{2013}+1}+\frac{48}{7^{2013}+1}=1\frac{48}{7^{2013}+1}\)
\(B=\frac{7^{2013}+1}{7^{2015}+1}\)
\(7^2.B=\)\(=\frac{7^{2015}+49}{7^{2015}+1}=\)\(\frac{7^{2015}+1+48}{7^{2015}+1}=\)\(\frac{7^{2015}+1}{7^{2015}+1}+\frac{48}{7^{2015}+1}=1\frac{48}{7^{2015}+1}\)
\(Vì\) \(1\frac{48}{7^{2013}+1}>1\frac{48}{7^{2013}+1}\)\(\Rightarrow7^2.A>7^2.B\)\(\Rightarrow A>B\)
\(Vậy\) \(A>B\)
Bài 2 nè
ta xét B trước:
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..\)\(.....+\frac{1}{2015}-\frac{1}{2016}\)
=\(\left(\frac{1}{1}+\frac{1}{3}+....+\frac{1}{2015}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{2016}\right)\)
\(=\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}\right)-\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
vậy A:B\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)\(:\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
\(=1\)
2A=2+22+23+...+22017
2A-A=22017-1
A=22017-1
=> A=B
mình ghi ứ ko hiểu thì hỏi nhé !
ta có :
2A = 2 + 2 mũ 2 + 2 mũ 3 ... + 2 mũ 2017
A= 2A - A = 2 mũ 2017 -1
=> A>B