Tính hợp lý
a.(1/2-1).(1/3-1).(1/4-1)...(1/2015-1)
b.(1/2^2-1).(1/3^2-1).(1/4^2-1)....(1/100^2-1)
Những câu hỏi liên quan
Chứng minh rằng:
a,A=1/2+1/2^2+1/2^3+.+1/2^2<1
b,B=1/3+1/3^2+1/3^3+...+1/3^n<1/2
c,B=1/2-1/2^2+1/2^3-1/2^4+...+1/2^2015-1/2^2016<1/3
d,D=1/3+2/3^2+3/3^3+4/3^4+...+100/3^100<3/4
?reeeeeeeeeeee
Ủa, cái số gì đây??????
Xem thêm câu trả lời
Bài 1 : Tính tổng
a) 1 *2 *3 + 2 * 3 *4 + 3 * 4 * 5 + ... + 2013 * 2014 * 2015 + 2014 * 2015 * 2016
b) 1 * + 3 * 4 + 5 * 6 + ... + 99 * 100
Bài 2 : CMR : 1^3 + 2^3 + 3^3 + ... + n^3 = ( 1 + 2 + 3 + ... + n )^2
BÀI 1:TÍNH:
\(B=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+....+\frac{100}{2^{100}}\)
BÀI 2: CHỨNG MINH RẰNG:
\(B=1-\frac{1}{2^2}-\frac{1}{3^2}-.....-\frac{1}{2004^2}>\frac{1}{2004}\)
BÀI 3:THỰC HIỆN PHÉP TÍNH BẰNG CÁCH HỢP LÝ:
\(B=\frac{1}{3}+\frac{1}{6}.\left(1+2\right)+\frac{1}{9}.\left(1+2+3\right)+.....+\frac{1}{6045}.\left(1+2+3+....+2015\right)\)
tính hợp lýa, A dfrac{1-dfrac{1}{sqrt{49}}+dfrac{1}{49}-dfrac{1}{left(7sqrt{7}right)^2}}{dfrac{sqrt{64}}{2}-dfrac{4}{7}+left(dfrac{2}{7}right)^2-dfrac{4}{343}}b, M 1 - dfrac{5}{sqrt{196}} - dfrac{5}{left(2sqrt{21}right)^2} - dfrac{sqrt{25}}{204} - dfrac{left(sqrt{5}right)^2}{374}
Đọc tiếp
tính hợp lý
a, A = \(\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}\)
b, M = 1 - \(\dfrac{5}{\sqrt{196}}\) - \(\dfrac{5}{\left(2\sqrt{21}\right)^2}\) - \(\dfrac{\sqrt{25}}{204}\) - \(\dfrac{\left(\sqrt{5}\right)^2}{374}\)
a: \(A=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4\left(1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}\right)}=\dfrac{1}{4}\)
b: \(M=1-\dfrac{5}{\sqrt{196}}-\dfrac{5}{\left(2\sqrt{21}\right)^2}-\dfrac{\sqrt{25}}{204}-\dfrac{\left(\sqrt{5}\right)^2}{374}\)
\(=1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)
\(=1-5\left(\dfrac{1}{14}+\dfrac{1}{84}+\dfrac{1}{204}+\dfrac{1}{374}\right)\)
\(=1-5\left(\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot12}+\dfrac{1}{12\cdot17}+\dfrac{1}{17\cdot22}\right)\)
\(=1-\left(\dfrac{5}{2\cdot7}+\dfrac{5}{7\cdot12}+\dfrac{5}{12\cdot17}+\dfrac{5}{17\cdot22}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)
\(=1-\dfrac{11-1}{22}=1-\dfrac{10}{22}=\dfrac{12}{22}=\dfrac{6}{11}\)
Đúng 1
Bình luận (0)
Bài 3: Tính một cách hợp lý:
a) A = 1 - 2 + 3 - 4 + .....+ 2013 - 2014 + 2015
b) 1 . 2 + 2 . 3 + 3 . 4 +....99 . 100
a) A = 1 - 2 + 3 - 4 + .....+ 2013 - 2014 + 2015
= (1 - 2) + (3 - 4) + .....+ (2013 - 2014) + 2015
=(-1) + (-1) + ...+ (-1) + 2015
1007 số -1
=(-1) . 1007 +2015
= -1007 + 2015
=1008
b) 1 . 2 + 2 . 3 + 3 . 4 +....99 . 100
đặt S= 1 . 2 + 2 . 3 + 3 . 4 +....99 . 100
=> 3S = 1.2.3 +2.3.(4-1) +...+99.100.(101-98)
= 1.2.3 +2.3.4-1.2.3+...+99.100.101-98.99.100
= 99.100.101
= 999900
=> S= 333300
Vậy 1 . 2 + 2 . 3 + 3 . 4 +....99 . 100 = 333300
Đúng 0
Bình luận (0)
a,A=(1-2)+(3-4)....(2013-2014)+2015
A= -1 + -1.....-1+2015
A= (2015-1):1+1
A=2015
A=(2015 x -1) x -1
A=2015
A=2015 + 2015
A=4030
b, 1/1.2 +1/2.3 ...1/99.100
1/1-1/2+1/2-1/3 ....1/99-1/100
1/1-1/100
99/100
Đúng 0
Bình luận (0)
Tính:
A=(1-1/1+2).(1-1/1+2+3).(1-1/1+2+3+4)...(1-1/1+2+3+4+...+2022)
B=1+1/2(1+2)+1/3(1+2+3)+1/100(1+2+3+...+100)
1. Tính hợp lí
a, A=11/125-17/18-5/7+4/9+17/14
b, B=1-1/2+2-2/3+3-3/4+4-1/4-3-1/3-2-1/2-1
c, C=1/100-1/100.99-1/99.98.......-1/3.2-1/2.1
khó nhìn lắm bn ak
sao pn ko cho
\(\frac{11}{125}-\frac{17}{18}-\frac{5}{8}+\frac{4}{9}+\frac{17}{14}.\)
thì có phải dễ nhìn hơn ko
Đúng 0
Bình luận (0)
a, \(A=\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)
\(=\frac{11}{125}+\left(\frac{-17}{18}+\frac{4}{9}\right)+\left(\frac{-5}{7}+\frac{17}{14}\right)\)
\(=\frac{11}{125}+\frac{-1}{2}+\frac{1}{2}\)
\(=\frac{11}{125}\)
b, \(B=1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)
\(=\left(1+2+3+4-3-2-1\right)-\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=4-3=1\)
c, \(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{-49}{50}\)
Đúng 0
Bình luận (0)
Cho
A = 1/2 + 1/3 + 1/4 + ... + 1/2017
B = 1/2016 + 2/2015 +3/2014+ ...+ 2015/2 + 2016/1
Tính B : A
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
Đúng 1
Bình luận (0)
Cho A = 1/2 + 1/3 + 1/4 + ... + 1/2017 B = 1/2015 + 2/2014 +3/2013 + ...+ 2015/2 + 2016/1 Tính B : A
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
Đúng 1
Bình luận (0)