So sánh
B=\(\dfrac{1}{2x4}\)+\(\dfrac{1}{4x6}\)+.....+\(\dfrac{1}{46x48}\) với \(\dfrac{1}{4}\)
HAKED BY PAKISTAN 2011
.jpg)
Những câu hỏi liên quan
So sánh:Adfrac{1}{1x2}+dfrac{1}{2x3}+.....+dfrac{1}{49x50} với 1Bdfrac{1}{2x4}+dfrac{1}{4x6}+.....+dfrac{1}{46x48} với dfrac{1}{4}Cdfrac{1}{6}+dfrac{1}{12}+.....+dfrac{1}{110}+dfrac{1}{122} với dfrac{1}{2}Giải tất cả các câu đầy đủ phép tính giúp mik nha. C.ơn
Đọc tiếp
So sánh:
A=\(\dfrac{1}{1x2}\)+\(\dfrac{1}{2x3}\)+.....+\(\dfrac{1}{49x50}\) với 1
B=\(\dfrac{1}{2x4}\)+\(\dfrac{1}{4x6}\)+.....+\(\dfrac{1}{46x48}\) với \(\dfrac{1}{4}\)
C=\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+.....+\(\dfrac{1}{110}\)+\(\dfrac{1}{122}\) với \(\dfrac{1}{2}\)
Giải tất cả các câu đầy đủ phép tính giúp mik nha. C.ơn
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\dfrac{49}{50}< 1\)
Đúng 0
Bình luận (1)
giúp tính \(\dfrac{2}{2x4}+\dfrac{2}{4x6}x...x\dfrac{2}{2014x2016}\)
Nguyễn Huy Tú soyeon_Tiểubàng giải Akai Haruma
Đúng 0
Bình luận (0)
Cho A= 1 + \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{4034}\); B = 1 + \(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{4033}\)
So sánh \(\dfrac{A}{B}\)với 1\(\dfrac{2017}{2018}\)
Cho \(A=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{4026}\)và \(B=1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{4025}\)So sánh với \(1\dfrac{2013}{2014}\)
Bạn thiếu đề rồi phải là trừ hay cộng j j chứ.
Xét:
`A+B=2+1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025`
`1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025>0`
`=>A+B>2`
Mà `1 2013/2014<2`
`=>A+B>1 2013/2014`
Đúng 3
Bình luận (0)
So sánh \(A\) với \(\dfrac{3}{4}\), biết \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
1/32< 1/2.3
1/42< 1/3.4
...
1/1002< 1/99.100
=> 1/22 + 1/32 + 1/42 + ... + 1/1002< 1/22 + 1/2.3 + 1/3.4 + ... + 1/99.100
A < 1/4 + 1/2 -1/3 + 1/3 - 1/4 +... + 1/99 - 1/100
A < 1/4 + 1/2 -1/100 < 1/4 + 1/2 = 3/4
=> A < 3/4
Đúng 1
Bình luận (0)
16 tháng 2 2022 lúc 21:46
so sánh với 1 :
\(\dfrac{1}{4444};\dfrac{3}{7};\dfrac{9}{5};\dfrac{7}{3};\dfrac{14}{15};\dfrac{16}{16};\dfrac{14}{11}\)
↑ \(\dfrac{1}{4}\) :>
\(\dfrac{1}{4444}< 1,\dfrac{3}{7}< 1,\dfrac{9}{5}>1,\dfrac{7}{3}>1,\dfrac{14}{15}< 1,\dfrac{16}{16}=1,\dfrac{14}{11}>1\)
Đúng 2
Bình luận (10)
1/4 < 1
3/7 < 1
9/5 > 1
7/3 > 1
14/15 < 1
16/16 = 1
14/11 >1
Đúng 2
Bình luận (0)
Cho M = \(1-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-\dfrac{1}{2^4}-....-\dfrac{1}{2^{10}}\) . So sánh M với \(\dfrac{1}{2^{11}}\)
\(M=1-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)
Đặt \(N=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)
\(2N=1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2N-N=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow N=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow M=1-\left(1-\dfrac{1}{2^{10}}\right)=\dfrac{1}{2^{10}}>\dfrac{1}{2^{11}}\)
Vậy \(M>\dfrac{1}{2^{11}}\)
Đúng 2
Bình luận (1)
cho A =(\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2024^2}\))
So sánh A với \(\dfrac{1}{2}\)
so sánh A=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{6^2}\)+...\(\dfrac{1}{\left(2n\right)^2}\)với \(\dfrac{1}{2}\)
A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+.......+\(\dfrac{1}{\left(2.n\right)^2}\)
A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{\left(2.2\right)^2}\)+ \(\dfrac{1}{\left(2.3\right)^2}\) +....+\(\dfrac{1}{\left(2.n\right)^2}\)
A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^2.2^2}\) + \(\dfrac{1}{2^2.3^2}\)+......+ \(\dfrac{1}{2^2.n^2}\)
A = \(\dfrac{1}{2^2}\) \(\times\) ( 1 + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+.......+ \(\dfrac{1}{n^2}\))
22 \(\times\) A = 1 + \(\dfrac{1}{2^2}\)+ \(\dfrac{1}{3^2}\)+......+\(\dfrac{1}{n^2}\)
4A = 1 + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) +......+ \(\dfrac{1}{n^2}\)
4A = 1 + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ...+\(\dfrac{1}{n.n}\)
1 = 1
\(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\)
\(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\)
...................
\(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right).n}\)
Cộng vế với vế ta có:
4A = 1+\(\dfrac{1}{2.2}\)+\(\dfrac{1}{3.3}\)+....+\(\dfrac{1}{n.n}\) <1+ \(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\)+ ......+ \(\dfrac{1}{\left(n-1\right).n}\)
4A < 1+ \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+....+\(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\) = 2 - \(\dfrac{1}{n}\)
A < ( 2 - \(\dfrac{1}{n}\)): 4
A < 2 : 4 - \(\dfrac{1}{n}\) : 4
A < \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\)
Vậy A < \(\dfrac{1}{2}\)
Đúng 2
Bình luận (0)
Ta có :22A=1+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{n^2}\)
22A-A=1-\(\dfrac{1}{\left(2n\right)^2}\)
3A=\(\dfrac{\left(2n\right)^2-1}{\left(2n\right)^2}\) <\(\dfrac{n^2}{\left(2n\right)^2}\)=\(\dfrac{1}{2}\)
3A<\(\dfrac{1}{2}\) suy ra A<\(\dfrac{1}{2}\)
Đúng 1
Bình luận (0)