so sanh a=(1/2-1)(1/3-1)(1/4-1)...(1/10-1)voi -1/9
Khong lam phep tinh hay so sanh:
a.[-1]*[-2]*[-3]*...*[-2009] voi 0
b.[-1]*[-2]*[-3]*...*[-10] voi 1*2*3*...*9*10
so sanh 1/4+1/9+1/10+1/41+1/42 voi 1/2
So sanh A voi 1:
A=1/2*2 + 1/3*3 + 1/4*4 + .....+1/2011*2011
So sanh B voi 3/4:
B=1/2*2 + 1/3*3 +1/4*4 + ......+1/2011*2011
cho A=1/4+1/9+1/16+1/25+...+1/10000 so sanh A voi 3/4
so sanh : 2001/2004 va 39/40
so sanh : A = 1/6 + 1/7 + 1/8 + 1/9 + 1/10 voi B = 1
\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{5}{5}=1=B\)
a/
\(\frac{2001}{2004}=\frac{2004-3}{2004}=1-\frac{3}{2004}=1-\frac{1}{668}.\)
\(\frac{39}{40}=\frac{40-1}{40}=1-\frac{1}{40}\)
Ta có \(40< 668\Rightarrow\frac{1}{40}>\frac{1}{668}\Rightarrow1-\frac{1}{40}< 1-\frac{1}{668}\Rightarrow\frac{39}{40}< \frac{2001}{2004}\)
b/
\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=1=B\)
A=(1/4-1).(1/9-1).(1/16-1)......(1/400-1) . so sanh A voi -1/2
Cho A=1/2!+2/3!+......+8/9!+9/10!. so sanh A voi 1.Voi n!=1.2.3.....n(tích của n số tự nhiên khác 0 đầu tiên)
A=\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{10}-1\right)\)
So sanh A voi \(-\frac{1}{9}\)
-A =( 1- 1/2 )(1 -1/3).....(1 -1/10)
= 1/2 . 2/3 ..... 9/10
= 1/10
-A = 1/10 nên A = -1/10
Vì 1/10 < 1/9 nên -1/10 > -1/9
Vậy A > -1/9
\(A=\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{10}-1\right)=-\frac{1}{2}.-\frac{2}{3}...-\frac{9}{10}\)
\(=\frac{-\left(1.2...9\right)}{2.3...10}=\frac{-1}{10}\)
(1/4-1).(1/9-1).....91/400-1) so sanh voi -1/2