so sanh A voi 1 biet A= 2^2019-(2^2018+2^2017+...+2^1+2^0)
Những câu hỏi liên quan
so sanh A B biet.
A=2017×2018-1/2017×2018
B=2018×2019-1/2018×2019
ban nao tra loi dung minh tich cho khong duoc lam tat
\(A=\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)(1)
\(B=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)(2)
Từ(1) và (2)
\(\Rightarrow B>A\)
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a. So sanh 2 phan so:A= 2015/2016+2016/2017+2017/2018 va B = 2015+2016+2017/2016+2017+2018
b.1/2.4+1/4.6+........+1/(2x-2).2x = 1/8
c.Cho A = 1/4+1/9+1/16+...+1/81+1/100 . Chung minh rang : A > 65/132
d.Cho B = 12/(2 . 4 ) ^ 2 + 20/ (4 . 6) ^2 + ...........+ 388/ ( 96 . 98 ) ^ 2 + 396/ ( 98 . 100 ) ^2 .Hay so sanh B voi 1 /4
So sánh A với 1 biết A= \(2^{2019}-\left(2^{2018}+2^{2017}+........+2^1+2^0\right)\)
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
So sanh giua 2*2019-1 voi 202099*2018
Giup minh nhanh nhe!
nhìn là biết rồi,cái bên trái
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)]
So sánh A=2^2019-2^2018-2^2017-.......-2-1 với 1
\(A=2^{2019}-2^{2018}-2^{2017}-...-2-1\)
\(A=2^{2019}-\left(2^{2018}+2^{2017}+...+2+1\right)=2^{2019}-B\)
Xét \(B=2^{2018}+2^{2017}+...+2+1\)
\(\Rightarrow2B=2^{2019}+2^{2018}+...+2^2+2\)
\(\Rightarrow2B-2^{2019}+1=2^{2018}+2^{2017}+...+2+1\)
\(\Rightarrow2B-2^{2019}+1=B\)
\(\Rightarrow B=2^{2019}-1\)
\(\Rightarrow A=2^{2019}-B=2^{2019}-\left(2^{2019}-1\right)=2^{2019}-2^{2019}+1=1\)
Vậy \(A=1\)
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cho a=1/1^3+1/3^2+...+1/3^2019 So sanh A voi 3/2
\(3a=3+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)
\(2a=3a-a=3-\frac{1}{3}-\frac{1}{3^{2019}}< 3\Rightarrow a< \frac{3}{2}\)