Cho M=\(\dfrac{1}{5}\)+\(\dfrac{2}{5^2}\)+\(\dfrac{3}{5^3}\)+...+\(\dfrac{2014}{5^{2014}}\). So sánh M với \(\dfrac{5}{36}\)
Những câu hỏi liên quan
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)
cho A=\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\)
So sánh A với\(\dfrac{1}{4}\)
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2015}}\)
\(\Rightarrow5A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{2014}}\)
\(\Rightarrow5A-A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{2014}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow4A=1-\dfrac{1}{5^{2015}}\)
\(\Rightarrow A=\dfrac{1}{4}-\dfrac{1}{5^{2015}.4}< \dfrac{1}{4}\)
\(\Rightarrowđpcm\)
Đúng 0
Bình luận (0)
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\)
\(\Rightarrow5A=5\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow5A=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\)
\(\Rightarrow5A-A=\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow4A=1-\dfrac{1}{5^{2015}}\)
\(\Rightarrow A=\dfrac{1}{4}-\dfrac{1}{5^{2015}.4}\)
\(\Rightarrow A< \dfrac{1}{4}\)
Đúng 0
Bình luận (0)
[Phan Đức Gia Linh _ Xin cảm ơn những người đã quan tâm tới câu hỏi của mình!]
a) Cho:
A 40 + dfrac{3}{8} + dfrac{7}{8^2} + dfrac{5}{8^3} + dfrac{32}{8^5}
B dfrac{24}{8^2} + 40 + dfrac{5}{8^2} + dfrac{40}{8^4} + dfrac{5}{8^4}
Hãy so sánh A và B
b) So sánh dfrac{2014^{2014}+1}{2014^{2015}+1} và dfrac{2014^{2013}+1}{2014^{2014}+1}
Các bạn cố gắng giải đầy đủ nhất giúp mình. Cảm ơn rất nhiều!
Đọc tiếp
[Phan Đức Gia Linh _ Xin cảm ơn những người đã quan tâm tới câu hỏi của mình!]
a) Cho:
A = 40 + \(\dfrac{3}{8}\) + \(\dfrac{7}{8^2}\) + \(\dfrac{5}{8^3}\) + \(\dfrac{32}{8^5}\)
B = \(\dfrac{24}{8^2}\) + 40 + \(\dfrac{5}{8^2}\) + \(\dfrac{40}{8^4}\) + \(\dfrac{5}{8^4}\)
Hãy so sánh A và B
b) So sánh \(\dfrac{2014^{2014}+1}{2014^{2015}+1}\) và \(\dfrac{2014^{2013}+1}{2014^{2014}+1}\)
Các bạn cố gắng giải đầy đủ nhất giúp mình. Cảm ơn rất nhiều!
Xàm như đọc bản tuyên ngôn không bằng
Đúng 0
Bình luận (1)
Xem thêm câu trả lời
16 tháng 7 2021 lúc 13:27
Bài 6: So sánha,dfrac{1}{2}+dfrac{1}{_{ }2^2}+dfrac{1}{2_{ }^3}+...+dfrac{1}{2^{2014}}và 1 b,dfrac{10^{2018}+5}{10^{2018}-8}và dfrac{10^{2019}+5}{10^{2019}-8}c,dfrac{1}{1.2.3}+dfrac{1}{2.3.4}+dfrac{1}{3.4.5}+...+dfrac{1}{23.24.25}vàdfrac{1}{4}
Đọc tiếp
Bài 6: So sánh
a,\(\dfrac{1}{2}\)+\(\dfrac{1}{_{ }2^2}\)+\(\dfrac{1}{2_{ }^3}\)+...+\(\dfrac{1}{2^{2014}}\)và 1 b,\(\dfrac{10^{2018}+5}{10^{2018}-8}\)và \(\dfrac{10^{2019}+5}{10^{2019}-8}\)
c,\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{23.24.25}\)và\(\dfrac{1}{4}\)
Cho B=\(\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{2014}{5^{2015}}\). Chứng tỏ rằng B<\(\dfrac{1}{16}\)
Cho:
A=\(\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
B=\(\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}\)
Tính \(\dfrac{B}{A}\)
giúp mình nha, mình đang cần gấp gấp gấp lắm....!HELP ME!!!
\(B=\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}\)
\(B=2016+\dfrac{2015}{2}+\dfrac{2014}{3}+....+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}\)
\(B=1+\left(\dfrac{2015}{2}+1\right)+\left(\dfrac{2014}{3}+1\right)+...+\left(\dfrac{3}{2014}+1\right)+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)\)
\(B=\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+....+\dfrac{2017}{2014}+\dfrac{2017}{2015}+\dfrac{2017}{2016}\)
\(B=2017\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
\(\dfrac{B}{A}=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}=2017\)
Đúng 0
Bình luận (0)
\(\dfrac{B}{A}=\dfrac{\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(\dfrac{2015}{2}+1\right)+\left(\dfrac{2014}{3}+1\right)+...+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\left(\dfrac{2015}{2}+\dfrac{2}{2}\right)+\left(\dfrac{2014}{3}+\dfrac{3}{3}\right)+...+\left(\dfrac{1}{2016}+\dfrac{2016}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
Vậy \(\dfrac{B}{A}=2017\)
Đúng 0
Bình luận (0)
1)so sánh 2 số sau M=\(\sqrt{18}-\sqrt{8}\) và N=\(\dfrac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}\)
2)cho biểu thức A=\((\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}):(\dfrac{x-4}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}})\) với x>0,\(x\ne4\),\(x\ne9\)
câu 2 rút gọn A và tìm các giá trị nguyên của x để A nhận giá trị âm
Đúng 0
Bình luận (0)
1) So sánh:
N = \(\dfrac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-\left(\sqrt{5}-1\right)=1\)
M = \(\sqrt{18}-\sqrt{8}\)
\(=3\sqrt{2}-2\sqrt{2}\)
\(=\sqrt{2}\)
Ta có: \(1=\sqrt{1}\)
Mà 1 < 2
\(\Rightarrow\sqrt{1}< \sqrt{2}\)
Hay 1 \(< \sqrt{2}\)
Vậy N < M
Đúng 1
Bình luận (0)
2) Với \(x>0;x\ne4;x\ne9\), ta có:
A = \(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{x-4}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(=\dfrac{x-3\sqrt{x}-2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-4-2\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x-3}\right)}\)
\(=\dfrac{-x-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-2\sqrt{x}+2}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-2\sqrt{x}+2}\)
\(=\dfrac{-x}{x-2\sqrt{x}+2}\)
Đúng 1
Bình luận (0)
Cho A = \(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}\). Chứng minh A < \(\dfrac{1}{4}\)
Help me!
A=\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}\)
5A=\(\dfrac{5}{5}+\dfrac{5}{5^2}+\dfrac{5}{5^3}+...+\dfrac{5}{5^{2014}}\)
5A=\(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}\)
5A-A=\(\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}\right)\)4A=\(1-\dfrac{1}{5^{2014}}\)
4A=\(\dfrac{5^{2014}-1}{5^{2014}}\)
A=\(\dfrac{5^{2014}-1}{5^{2014}}:4\)
A=\(\dfrac{5^{2014}-1}{5^{2014}}.\dfrac{1}{4}\)
\(\Rightarrow\)A<\(\dfrac{1}{4}\)
Đúng 0
Bình luận (0)
Ta có:
A = \(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{2014}}\)
\(\Rightarrow\) 5A = 5\(\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{2014}}\right)\)
\(\Rightarrow\) 5A = \(\dfrac{5}{5}+\dfrac{5}{5^2}+\dfrac{5}{5^3}+....+\dfrac{5}{5^{2014}}\)
\(\Rightarrow\) 5A = \(1+\dfrac{1}{5}+\dfrac{1}{5^2}+....+\dfrac{1}{5^{2013}}\)
\(\Rightarrow\)\(\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+....+\dfrac{1}{5^{2013}}\right)\)-\(\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{2014}}\right)\) = 5A - A
\(\Rightarrow\)4A= 1 - \(\dfrac{1}{5^{2014}}\)
\(\Rightarrow\) A =\(\dfrac{5^{2014}-1}{5^{2014}}\) : 4
Vậy A =\(\dfrac{5^{2014}-1}{5^{2014}}\) : 4
Đúng 0
Bình luận (0)
25 tháng 7 2023 lúc 10:18
Bài 1: Tìm x; y ϵ ℤ
a) 2x - ysqrt{6} 5 + (x + 1)sqrt{6}
b) 5x + y - (2x -1)sqrt{7} ysqrt{7} + 2
Bài 2: So sánh M và N
M dfrac{dfrac{3}{4}+dfrac{3}{5}+dfrac{3}{7}-dfrac{3}{11}}{dfrac{6}{4}+dfrac{6}{5}+dfrac{6}{7}-dfrac{6}{11}}
N dfrac{dfrac{2}{3}+dfrac{2}{5}-dfrac{2}{7}-dfrac{2}{11}}{dfrac{6}{2}+dfrac{6}{5}-dfrac{6}{7}-dfrac{6}{11}}
Bài 3: Chứng minh:
dfrac{1}{2!}+dfrac{1}{3!}+dfrac{1}{4!}+...+dfrac{1}{2023!} 1
Đọc tiếp
Bài 1: Tìm x; y ϵ \(ℤ\)
a) 2x - y\(\sqrt{6}\) = 5 + (x + 1)\(\sqrt{6}\)
b) 5x + y - (2x -1)\(\sqrt{7}\) = y\(\sqrt{7}\) + 2
Bài 2: So sánh M và N
M = \(\dfrac{\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{6}{4}+\dfrac{6}{5}+\dfrac{6}{7}-\dfrac{6}{11}}\)
N = \(\dfrac{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}{\dfrac{6}{2}+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}\)
Bài 3: Chứng minh:
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
Đúng 1
Bình luận (0)