Tính
a,A=(1/4-1)*(1/4-1)*(1/9-1)*(1/16-1)*.....*(1/100-1)*(1/121-1)
b,S=2^2010-2^2019-2^2018-.....-2-1
Câu 1:
a, Tính A= (-1)*(-1)^2*(-1)^3*(-1)^4.....(-1)^2018*(-1)^2019.
b, Tính tổng S= 1*2*3+2*3*4+...+2018*2019*2020
A = (-1)(-1)^2(-1)^3...(-1)^2019
A = (-1)^1+2+3+...+2019
A = (-1)^2039190
A = 1
S = 1.2.3 + 2.3.4 + 3.4.5 + ... + 2018.2019.2020
4S = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + .... + 2018.2019.2020.4
4S = 1.2.3.4 + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) + ... + 2018.2019.2020.(2021 - 2017)
4S = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 2018.2019.2020.2021 - 2017.2018.2019
4S = 2018.2019.2020.2021
S = 2018.2019.2020.2021 : 4 = ...
Bài 1:
a, Tính A= \(\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right).\left(\dfrac{1}{16}-1\right).....\left(\dfrac{1}{100}-1\right).\left(\dfrac{1}{121}-1\right)\)
b, S=\(2^{2010}-2^{2009}-2^{2008}-......-2-1\)
a,
\(\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}........\dfrac{-99}{100}.\dfrac{-120}{121}\)
\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.........\dfrac{9.11}{10^2}.\dfrac{10.12}{11^2}\)
\(=\dfrac{1.2.3.4.....10.3.4.5.6......11.12}{2^2.3^2........11^2}\)
\(=\dfrac{1.2.11.12}{2^2.11^2}=\dfrac{12}{22}\)
\(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\\ \Rightarrow S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)
Đặt \(M=2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow S=2^{2010}-M\)
* Tính M
\(M=2^{2009}+2^{2008}+...+2+1\\ \Rightarrow2^0+2^1+...+2^{2008}+2^{2009}\\ \Rightarrow2S=2^1+2^2+...+2^{2009}+2^{2010}\\ \Rightarrow2S-S=\left(2^1+2^2+...+2^{2009}+2^{2010}\right)-\left(2^0+2^1+...+2^{2008}+2^{2009}\right)\\ \Rightarrow S=2^{2010}-2^0=2^{2010}-1\)Thay M vào S, ta được :
\(S=2^{2010}-\left(2^{2010}-1\right)\\ \Rightarrow S=2^{2010}-2^{2010}+1\\ \Rightarrow S=1\)
Tính A:B
a)A=98+1/2+1/3+1/4+...+1/99
B=2/3+4/3+5/4+...+100/99
b)A=2018+1/2+1/3+1/4+...+1/2019
B=3/2+4/3+3/4+...+2020/2019
c)A=99/1+98/2+97/3+...+2/98+1/99
B=1/2+1/3+1/4+...+1/100
Giải đầy đủ
a) \(A=98+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}=1\)
b) \(A=2018+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\)(có 2018 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{2019}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}=1\)
c) \(A=\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}\)
\(A=99+\frac{98}{2}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào từng phân số)
\(A=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+1\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Và \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\)
\(\Rightarrow\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}}=100\)
a)\(B=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+...+\frac{100}{99}\)
\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{99}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\right)\)
\(\Rightarrow B=98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
b)\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{2019}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right)\)
\(\Rightarrow B=2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}=1.\)
Vậy \(A:B=1.\)
c)\(A=\left(1+1+...+1\right)+\frac{98}{2}+\frac{97}{3}+...+\frac{2}{98}+\frac{1}{99}\)
\(A=\left(1+\frac{98}{2}\right)+\left(1+\frac{97}{3}\right)+...+\left(1+\frac{2}{98}\right)+\left(1+\frac{1}{99}\right)\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{98}+\frac{100}{99}\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
Cho A=1/2+1/3+1/4+...+1/2019 và B=2018/1+2017/2+...+2/2017+1/2018. So sánh A/B với 1/2018
Cho: A=1-1/2+1/3-1/4+...-1/2018+1/2019
B=1/1010+1/1011+...+1/2018+1/2019
Tính (A-B)2020
a,tính:A=\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right).....\left(\frac{1}{100}-1\right).\left(\frac{1}{121}-1\right)\)
b,tình=\(2^{2010}-2^{2009}-2^{2008}-...-2-1\)
giúp với!!!!!♫♪
cho A= 1/2 + 1/3 + 1/4 +...+ 1/2020 và B= 2019/1 + 2018/2 + 2017/3 +...+ 1/2019. Tính A/B
Tính tỉ số A/B biết:
A=1/2 + 1/3 + 1/4 + ... + 1/2017 + 1/2018 + 1/2019
B=2018/1 + 2017/2 + 2016/3 + ... + 2/2017 + 1/2018
\( S =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1} {2019}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right) \)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(\(\Rightarrow S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\) \(\Rightarrow S=P\)\)
\(B=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)
\(B=1+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{1}{2018}+1\right)\)
\(B=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)
\(B=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)
ta có \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}=\frac{1}{2019}\)
Cho A=1/2018+2/2017+3/2016+...+2017/2+2018
B=1/2+1/3+1/4+....+1/2019
Tính A/B
\(A=\frac{1}{2018}+\frac{2}{2017}+...+\frac{2017}{2}+2018\)
\(=\left(\frac{1}{2018}+1\right)+\left(1+\frac{2}{2017}\right)+...+\left(\frac{2017}{2}+1\right)+1\)(2018 số hạng 1)
\(=\frac{2019}{2018}+\frac{2019}{2017}+...+\frac{2019}{2}+\frac{2019}{2019}=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)\)
Mà \(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
=> Khi đó : \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)
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