so sánh:
1/1^2+1/3^2+1/4^2+.....+1/2018^2 với 1
so sánh A=1/2^2+1/4^2+1/6ˆ2+...+1/2018^2 với 1/2
so sánh 1-1/2+1/3-1/4+....-1/2018 với 1/1005+1/1006+.......+1/2008
So sánh bt :1/1!+1/2!+1/3!+1/4!+1/5!+...+1/2018! với 2017/1152
A=1/2^2+1/3^2+1/4^2+.....+1/2018^2 với B=75%
So sánh A và B
\(\frac{1}{3^2}< \frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3},\frac{1}{4^2}< \frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4},...,\frac{1}{2018^2}< \frac{1}{2017\cdot2018}=\frac{1}{2017}-\frac{1}{2018}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}=\frac{1}{2}-\frac{1}{2018}< \frac{1}{2}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}< \frac{1}{2^2}+\frac{1}{2}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}=75\%\)
so sánh 1+3+3 mũ 2 +...+3 mũ 2017 với 3 mũ 2018-1 trên 2 với 3 mũ 2018 trên 2
Cho A=1+2+2^2+2^3+2^4+...2^2017. So sánh A với 2^2018
cho A = 1/2^2 + 1/3^2 + 1/4^2 + ...+ 1/2018^2 và B = 75%. So sánh A và B
= 1/2.2 + 1/3.3 + ... + 1/2018.2018
= ( 1/2 - 1/2) + (1/3 - 1/3) + ... + ( 1/2018 - 1/2018 )
= 0+0+0+0+...+0
=0
75% = 7,5
7,5 > 0 ==>
A<B
Cho A=1/2^2+1/3^2+1/4^2+...+1/2018^2 và B=75%. So sánh A và B
B = 75% => B = 3/4
Ta có :\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}=1-\frac{1}{2018}\)
Vì \(\frac{1}{2018}< \frac{1}{4}\Rightarrow1-\frac{1}{2018}>1-\frac{1}{4}\Rightarrow A>\frac{3}{4}\)=> A > B
\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}\)
\(B=75\%=\frac{3}{4}\)
Ta có:\(A=.......\)
\(=\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\right)< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2018}=\frac{3}{4}-\frac{1}{2018}< \frac{3}{4}\)
\(\Rightarrow A< B\)
cho A=\(\frac{1+2+2^2+2^3+2^4+...+2^{2018}}{2^{2019}}\) so sánh A với 1
2A = 2+ 2^2+.......+2^2018+ 2^2019 / 2 ^2019
2A-A = 2+2^2+.....+2^2019/ 2^2019 - 1=2+....+2^2018/ 2^2019
A =2^2019-1 / 2^2019 < 1
suy A< 1
xin lỗi mình hơi nhầm : 2A - A = 2+2^2+....+2^2019/2^2019- 1+2+2^2 +.....+2^2019/2^2019
\(A=1+2+2^2+2^3+2^4+..+2^{2018}\)
\(2A=2+2^2+2^3+...+2^{2019}\)
\(2A-A=\left(2+2^2+2^3+...+2^{2019}\right)-\left(1+2+2^2+..+2^{2018}\right)\)
\(A=\left(2^{2019}-1\right):2019>1\)