2^2019 - 2^2018 - 2^2017 - ... - 2^3 - 2^2 - 2 - 1
Những câu hỏi liên quan
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/1)+(2019/2)+(2019/3)+(2019/4)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/2)+(2019/3)+(2019/4)+(2019/5)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
Bài 1:so sánh: 2017/2018+2018/2019 và ( 2017+2018/2018/2019)
Bài 2: (1/2003+1/2004+1/2005)/(2/2003+2/2004+2/2005)
Bài 3: 2013+ (2013/1+2)+(2013/1+2+30+...+(2013/1+2+3+..+2012)
Bài 1
\(\frac{2017}{2018}+\frac{2018}{2019}\)và \(\left(\frac{2017+2018}{2018+2019}\right)\)mk chữa lại đề luôn đó
Ta tách :
\(\frac{2017}{\left(2018+2019\right)+2018}\)
đến đây ta tách
\(\frac{2017}{2018+2019}< \frac{2017}{2018}\)
vậy....
mấy câu khác tương tự
Đúng 0
Bình luận (0)
2) \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{\frac{2}{2003}+\frac{2}{2004}+\frac{2}{2005}}\)
= \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{2.\frac{1}{2003}+2.\frac{1}{2004}+2.\frac{1}{2005}}\)
=\(\frac{1\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}{2.\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}\)
= \(\frac{1}{2}\)
3) \(2013+\left(\frac{2013}{1+2}\right)+\left(\frac{2013}{1+2+3}\right)+...+\left(\frac{2013}{1+2+3+...+2012}\right)\)
= \(2013.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}\right)\)
= \(2013.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{2025078}\right)\)
= \(2013.2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4050156}\right)\)
=\(4026.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)\)
= \(4026.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
= \(4026.\left(1-\frac{1}{2013}\right)\)
= \(4026.\frac{2012}{2013}\)
=\(4024\)
Đúng 0
Bình luận (0)
cho A =1+2^2018+3^2017+4^2016+...+2018^2+2019,B=1+2^2017+3^2016+...+2017^2+2018,chứng tỏ giá trị biểu thức A-3B dương
cho A =1+2^2018+3^2017+4^2016+...+2018^2+2019,B=1+2^2017+3^2016+...+2017^2+2018,chứng tỏ giá trị biểu thức A-3B dương
1×2×3×4×...×2018×2019-1×2×3×...×2018-1×2×3×4×..×2017×20182
1x2x3x...2018x2019 - 1x2x3x..2018 - 1x2x3x4x...x2017x20182
= 1x2x3x...x2018x(2019 - 1 - 2018)
= 1x2x3x...x2018x0
= 0
Đúng 0
Bình luận (0)
Cho P =1/2×1/3+...+1/2018+1/2019 và Q = 1/2018+2/2017 +...+ 2017/2+2018/1
Tính (Q - P .2019)^2019
Các bạn giúp mình với nha. Cảm ơn rất nhiều !
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 \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}=\frac{1}{2019}\)
A=1/2+1/3+1/4+...+1/2019;B=1/2018 +2/2017+3/2016+...+2017/2+2018/1.Tính A/B
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 : \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)
??????????????????????????????????????????????????????????????????????????????????????????????????????????????