Tính B = 2014/1 + 2013/2 + 2013/3 + ... + 1/2014
NM
Những câu hỏi liên quan
So sánh ;A=2014^2013+1/2014^2013-1 va B=2014^2013-1/2014^2013-3.giup mk vs
ta có: \(A=\frac{2014^{2013}+1}{2014^{2013}-1}=\frac{2014^{2013}-1+2}{2014^{2013}-1}=1+\frac{2}{2014^{2013}-1}\)
\(B=\frac{2014^{2013}-1}{2014^{2013}-3}=\frac{2014^{2013}-3+2}{2014^{2013}-3}=1+\frac{2}{2014^{2013}-3}\)
\(\Rightarrow\frac{2}{2014^{2013}-1}< \frac{2}{2014^{2013}-3}\)
\(\Rightarrow1+\frac{2}{2014^{2013}-1}< 1+\frac{2}{2014^{2013}-3}\)
=> A < B
Đúng 0
Bình luận (0)
Cho:
A= 1/1*2+1/3*4+1/5*6+...+1/2013*2014
B=1/1008*2014+1/1009*2013+...+1/2014*1008
Tính A/B
Thực hiện tính :
a) A = 1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/2013(1+2+3+..+2013)
b) B = 1-3/7.3+2-4/2.4+3-5/3.5+4-6/4.6+....+2011-2013/2011.2013+2012-2014/2012.2014-2013+2014/2013.2014
(1/2012+1/2013-1/2014)/(5/2012+5/2013-5/2014)-(2/2103+2/2014-2/2015)/(3/2013+3/2014-3/2015)
\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)
=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)
Đúng 0
Bình luận (0)
Cho A dfrac{1}{2014}+dfrac{2}{2013}+dfrac{3}{2012}+...+dfrac{2013}{2}+2014 B dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+...+dfrac{1}{2015} Tính giá trị dfrac{A}{B}
Đọc tiếp
Cho A = \(\dfrac{1}{2014}\)+\(\dfrac{2}{2013}\)+\(\dfrac{3}{2012}\)+...+\(\dfrac{2013}{2}\)+2014
B = \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+...+\(\dfrac{1}{2015}\)
Tính giá trị \(\dfrac{A}{B}\)
A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B
\(\Rightarrow\) \(\dfrac{A}{B}\)=2015
Đúng 0
Bình luận (0)
1) 1/2 + 1/3 + 1/4 + ... + 1/2013 + 1/2014
2) 2014 + 2013/2 + 2012/3 + 2011/4 + ... + 2/2013 + 1/2014
H24
14 tháng 2 2020 lúc 10:57
Tính các tổng sau:
a) A=1+(-2) + 3 +(-4) + ...+(- 2014) + 2015;
b) B= (-2) + 4 +(-6) + 8 ... +(-2014) + 2016;
c) 1+(-3) + 5 +(-7) + ... + 2013 +(-2015);
d) (-2015) + (-2014) + (-2013)+ ... + 2015 + 2016
\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)
\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)
\(B=\left(-2\right)+4+\left(-6\right)+8+\left(-10\right)+,...+\left(-2014\right)+2016\)
\(B=2+2+....+2\left(\text{504 số hạng 2}\right)=1008\)
c) 1 + ( -3 ) +5 + ( -7 ) + ...........+ 2013 + ( -2015 )
[ 1 + (-3 ) ] + [ 5 + -7 ] + .......... + [ 2013 + ( - 2015 ) ]
có số cặp là : [ ( 2015 - 1 ) : 2 + 1 ] : 2 = 504 ( cặp )
= -2 + -2 + -2 +..........+ -2
= -2 x 504
= -1008
Xem thêm câu trả lời
Tính
A=3^2016 - 3^2015 + 3^2014 - 3^2013 + ......+ 3^2 - 3 + 1
B= 4^2016 - 4^2015 + 4^2014 - 4^2013 + ......+4^2 - 4 + 1
=>3A= 3^2017-3^2016+3^2015-...-3^2+3
=>3A+A=4A=3^2017+1=>A=\(\frac{3^{2017}+1}{4}\)
B tương tự nha
Đúng 0
Bình luận (0)
frac{frac{1}{2}+frac{1}{3}+......+frac{1}{2013}}{frac{2012}{1}+frac{2012}{2}+frac{2011}{3}+...+frac{1}{2013}}frac{frac{1}{2}+frac{1}{3}+..+frac{1}{2013}}{frac{2012}{1}+2+frac{2012}{2}+1+frac{2011}{3}+1+...+frac{1}{2013}+1-2014}frac{frac{1}{2}+frac{1}{3}+...+frac{1}{2013}}{frac{2014}{1}+frac{2014}{2}+...+frac{2014}{2013}-2014}frac{frac{1}{2}+frac{1}{3}+...+frac{1}{2013}}{2014left(1+frac{1}{2}+frac{1}{3}+...+frac{1}{2013}-1right)}frac{1}{2014}
Đọc tiếp
\(\frac{\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2013}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}}{\frac{2012}{1}+2+\frac{2012}{2}+1+\frac{2011}{3}+1+...+\frac{1}{2013}+1-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\frac{2014}{1}+\frac{2014}{2}+...+\frac{2014}{2013}-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2014\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-1\right)}\)
=\(\frac{1}{2014}\)