Tính hợp lí
1+1/5+3/35+1/21+1/33+...+1/3201+1/3333
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1: =(-20/36+15/36)*(-3/10)+(-16/36+21/36)
=-5/36*(-3/10)+5/36
=5/36*13/10
=65/360=13/72
2: Bạn xem lại đề nha bạn, dãy số này không có quy luật nào hết luôn á
Bài 1:
1: =15+37+52-37-17=52-2=50
2: =38-42+14-25+27+15=62-42+29=20+29=49
Bài 1: Bỏ ngoặc rồi tính
3) (21-32) - (-12+32)=21-32-(-12)-32=21-32+12-32=-31
4) (12+21) - (23-21+10)=12+21-23+21-10=21
5) (57-725) - (605-53)=57-725-605+53=-1220
6) (55+45+15) - (15-55+45)=55+45+15-15+55-45=55+55=110
Bài 2: Tính các tổng sau một cách hợp lí
1) (-37) + 14 + 26 + 37=(-37+37)+(14+26)=0+40=40
2) (-24) +6 + 10 + 24=(-24+24)+(6+10)=0+16=16
3) 15 + 23 + (-25) + (-23)=(15-25)+(23-23)=-10+0=-10
4) 60 + 33 + (-50) + (-33)=(60-50)+(33-33)=10+0=10
5) (-16) + (-209) + (-14) + 209=(-16-14)+(-209+209)=-30+0=-30
6) (-12) + (-13) + 36 + (-11)=(-11-12-13)+36=-36+36=0
A = ( 15/22 - 2/22 ) : 1/33 - ( 6/84 - 8/84) : 1/35 + 1 + 1/5 . ( 3/12 - 2/12 - 12/12 ) . 5/11
A = 13/22 . 33 - (-1/42) . 35 + 1 + 1/5 . - 11/12 . 5/11
A = 39/2 - ( -5/6 ) + 1 + - 11/60 .5/11
A = 39/2 + 5/6 + 1 + (- 1/12)
A = 234/12 + 10 /12 + 12/12 + (-1/12)
A = 255/12
1: \(\dfrac{4}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\)
\(=1+\dfrac{1}{2}\)
\(=\dfrac{3}{2}\)
2: \(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}\)
1. 8 - ( 2014 - 2008 ) + 2014
= 8-2014+2008+2014
= (8-2008)+(-2014+2014)
= -2000
2. 5679 + ( 1357 - 5679 - 17
= 5679+1357-5679-17
= 1340
3. 1268 - ( 78 + 1268 ) - ( -78 )
= 1268-78-1268+78
= 0
4. 13567 - ( 15 - 27 )
= 13567+12
= 13579
5. - 48795 - ( 489 - 48795 ) + 400
= -48795-489+48795+400
= (-48795+48795)+(-489+400)
= -89
6. 15641 - ( 27 + 15641 )
= 15641-27-15641
= -27
a) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{19}{20}=\frac{1}{20}\)
b) \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
=> \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
=> \(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
=> \(A=2-\frac{1}{2^{2012}}\)
c) \(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{231}{4}.\frac{4}{21}=11\)
d.e) ktra lại đề
Ta có :
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
Vậy \(A=\frac{25}{17}\)
Chúc bạn học tốt ~
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
\(B=\frac{21}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\frac{4}{21}\)
\(B=\left(\frac{21}{4}.\frac{4}{21}\right).33\)
\(B=33\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(C=\frac{1}{2}.\frac{98}{99}\)
\(C=\frac{49}{99}\)
tick minh minh tra loi cho
đề có vẻ ko ổn bạn xem lại