a- Cho A= 1+2+2^2+2^3+...+2^9 và B= 5*2^8. Hãy so sánh A và B
b- So sánh: A= 2/60*63+2/63*66+...+2/117*120+2/2003
B= 5/40*44+5/44*48+...+5/76*80+5/2003
So sánh A và B biết:
A= 2/60*63+2/63*66+.....+2/117.120+2/2003
B= 5/40*44+5/44*48+.....+5/76*80+5/2003
Ta co
+)A=2/60*63+2/63*66+...+2/117*120+2/2003
A*3/2=3/60*63+3/63*66+...+3/117*120+3/2003
A*3/2=1/60-1/63+1/63-1/66+...+1/117-1/120+3/2003
A*3/2=1/60-1/120+3/2003
A=(1/120+3/2003)*2/3
+)B=5/40*44+5/44*48+...+5/76*80+5/2003
B*4/5=4/40*44+4/44*48+...+4/76*80+4/2003
B*4/5=1/40-1/44+1/44-1/48+...+1/76-1/80+4/2003
B*4/5=1/40-1/80+4/2003
B=(1/80+4/2003)*5/4
Tu tren ta co A=(1/120+3/2003)*2/3
B=(1/80+4/2003)*5/4
Vay A<B(Vi 1/120<1/80;3/2003<4/2003;2/3<5/4)
+)A=2/60*63+2/63*66+...+2/117*120+2/2003
A*3/2=3/60*63+3/63*66+...+3/117*120+3/2003
A*3/2=1/60-1/63+1/63-1/66+...+1/117-1/120+3/2003
A*3/2=1/60-1/120+3/2003
A=(1/120+3/2003)*2/3
+)B=5/40*44+5/44*48+...+5/76*80+5/2003
B*4/5=4/40*44+4/44*48+...+4/76*80+4/2003
B*4/5=1/40-1/44+1/44-1/48+...+1/76-1/80+4/2003
B*4/5=1/40-1/80+4/2003
B=(1/80+4/2003)*5/4
Tu tren ta co A=(1/120+3/2003)*2/3
B=(1/80+4/2003)*5/4
Vay A<B(Vi 1/120<1/80;3/2003<4/2003;2/3<5/4)
C1:Ta co
+)A=2/60*63+2/63*66+...+2/117*120+2/2003
A*3/2=3/60*63+3/63*66+...+3/117*120+3/2003
A*3/2=1/60-1/63+1/63-1/66+...+1/117-1/120+3/2003
A*3/2=1/60-1/120+3/2003
A=(1/120+3/2003)*2/3
+)B=5/40*44+5/44*48+...+5/76*80+5/2003
B*4/5=4/40*44+4/44*48+...+4/76*80+4/2003
B*4/5=1/40-1/44+1/44-1/48+...+1/76-1/80+4/2003
B*4/5=1/40-1/80+4/2003
B=(1/80+4/2003)*5/4
Tu tren ta co A=(1/120+3/2003)*2/3
B=(1/80+4/2003)*5/4
Vay A<B(Vi 1/120<1/80;3/2003<4/2003;2/3<5/4)
C2:
b:
Vì 1/360+1/2003<1/320+1/2003
nên A<B
So sánh :
a= 2 / 60 x 63 + 2 / 63 x 66 + 2 / 66 x 69 + ........... + 2 / 117 x 120 + 2 / 2011
và b= 5 / 40x 44 + 5 / 44 x 48 + 5 / 48 x 52 + ......... + 5 / 76 x 80 + 5 / 2011
Làm theo cách này nhé :
a = 2 / 60 x 63 + 2 / 63 x 66 + 2 / 66x 69 + ...+ 2 / 117 x 120 + 2 / 2011
= 2/3 x ( 3/60 x 63 + 3 / 63 x 66 + 3 / 66 x 69 + ...+ 3/117 x 120 ) + 2/2011
= 2/3 x ( 1/60 - 1/63 + 1/63 - 1/66 + 1/66 - 1/69 + ... + 1/117 - 1/120 ) + 2/2011
= 2/3 x ( 1/60 - 1/120 ) + 2/2011
= 2/3 x 1/120 + 2/2011
= 1/180 + 2/2011
b = 5/ 40 x 44 + 5 / 44 x 48 + ...+ 5/76 x 80 + 5/ 2011
= 5/4 x ( 4/40 x 44 + 4/44 x 48 + ...+ 4/76 x 80 ) + 5/2011
= 5/4 x ( 1/40 - 1/44 + 1/44 - 1/48 + ...+ 1/76 - 1/80 ) + 5/2011
= 5/4 x ( 1/40 - 1/80 ) + 5/2011
= 5/4 x 1/80 + 5/2011
= 1/64 + 5/2011
Do 1/64 > 1/80 ; 5/2011 > 2/2011
=> 1/64 + 5/2011 > 1/80 + 2/2011
=> b > a
K nha
Mình sửa lại chút nhé , lỗi đánh bàn phím thoy , :
Do 1/64 > 1/180 ; 5/2011 > 2/2011
=> 1/64 + 5/2011 > 1/180 + 2/2011
=> b > a
so sánh:
A=2/60×63+2/63×66+...+2/114×117+2/2003
B=5/40×44+5/44×48+...+5/76×80+5/2003
C=124×(1/1×1985+1/2×1086+...+1/16×2000)
D=1/1×17+1/2×18+...+1/1984×2000
bài đội tuyển đấy
bạn nào làm đc mình tick cho
Pleas! help me!
so sánh:
A=2/60×63+2/63×66+...+2/114×117+2/2003
B=5/40×44+5/44×48+...+5/76×80+5/2003
C=124×(1/1×1985+1/2×1086+...+1/16×2000)
D=1/1×17+1/2×18+...+1/1984×2000
bạn nào làm đc mình tick cho
Pleas! help me!
So sánh A=\(\frac{2}{60\times63}+\frac{2}{63\times66}+...+\frac{2}{117\times120}+\frac{2}{2003}\)
B=\(\frac{5}{40\times44}+\frac{5}{44\times48}+...+\frac{5}{76\times80}+\frac{5}{2003}\)
Ta có: \(A=\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}\left(\frac{3}{60.63}+\frac{3}{63.66}+...+\frac{3}{117.120}\right)+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{120}\right)+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}.\frac{1}{120}+\frac{2}{2003}\)
\(\Rightarrow A=\frac{1}{180}+\frac{2}{2003}\)
\(B=\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}\left(\frac{4}{40.44}+\frac{4}{44.48}+...+\frac{4}{76.80}\right)+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}.\frac{1}{80}+\frac{5}{2003}\)
\(\Rightarrow B=\frac{1}{64}+\frac{5}{2003}\)
Vì \(\left\{\begin{matrix}\frac{1}{64}>\frac{1}{180}\\\frac{5}{2003}>\frac{2}{2003}\end{matrix}\right.\Rightarrow\frac{1}{64}+\frac{5}{2003}>\frac{1}{180}+\frac{2}{2003}\Rightarrow B>A\)
Vậy A < B
1,So sánh :
a, A=2/60*63+2/63*66+2/66*69+...+2/117*120+2/2011 với B=5/40*44+5/44*48+5/48*52+...+5/76*80+5/2011
b,Cho C=2222...220000...0077777...77.C là số nguyên tố hay hợp số. Biết có 2011 số 2, 2011 số 0, 2011 số 7
2,Cho D=1-2+3-4+.....+99-100
a, D có chia hết cho 2, cho 3, cho 5 không? Vì sao?
b,D có bao nhiêu ước tự nhiên ? Có bao nhiêu ước nguyên?
CỨU!!!!
Cho
\(A=\frac{2}{60\cdot63}+\frac{2}{61\cdot64}+...+\frac{2}{117\cdot120}+2011\)
\(B=\frac{5}{40\cdot44}+\frac{5}{44\cdot48}+...+\frac{5}{76\cdot80}+\frac{2}{2011}\)
Hãy so sánh A và B
1/Chứng minh : 5^2003+5^2002+5^2001 chia hết cho 31'
2./ Cho A = 1+2+2^2+......+2^9+2^10 và B = 2^11- 1 .So sánh A B
2/
A=1+2+2^2+...+2^10
2.A= 2+2^2+...+2^11
=>2A-A = 2^11-1=> A = 2^11 -1=B
Vậy A=B
1)52003+52002+52001=52001(52+5+1)=52001(25+5+1)=52001.31
Vì 31 chia hết cho 31nên
52001.31chia hết cho 31 hay 52003+52002+52001 chia hết cho 31
2) A = 1+2+22+......+29+210
=>2A=2+22+23+...+211
=>2A-A=2+22+23+...+211-(1+2+22+...+29+210)
=>A=211-1
Vậy A=B=211-1
Bài 1: so sánh
a, 2711 và 818
b, 536 và 1124
d, 1520 và 910.521
e, Cho A= 3+ 22 +23+24+...........+22002 + 22003 và B= 22003. So sánh A và B
f,4+ 32 + 33+...............32004 C= 32005. So sánh A và B