1) Tìm 2 số nguyên tố x, y sao cho: \(x^2-6y^2=1\)
2) Cho \(B=1.2.3...2012.\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)\)
CMR: B chia hết cho 2013
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7 tháng 4 2017
c) Cho B = (1.2.3....2012) . ( 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2012}\) ) Chứng minh B chia hết cho 2013
B = (1.2.3....2012) . (1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ...+ \(\dfrac{1}{2012}\) )
=(1.2.3...671...2012) . (1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2012}\))
=(1.2.(3.671)...2012) . (1 + \(\dfrac{1}{2}\) +\(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2012}\))
=(1.2.2013...2012) . (1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2012}\))
Vậy B chia hết cho 2013
Đúng đấy, bạn cứ chép vào đi
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 11 2021
a.
\(a^2+a^2\left(a+1\right)^2+\left(a+1\right)^2=a^2+\left(a^2+a\right)^2+a^2+2a+1\)
\(=\left(a^2+a\right)^2+2\left(a^2+a\right)+1=\left(a^2+a+1\right)^2\)
b.
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{y}\right)^2-\dfrac{x}{y}=3\\x+\dfrac{1}{y}+\dfrac{x}{y}=3\end{matrix}\right.\)
\(\Rightarrow\left(x+\dfrac{1}{y}\right)^2+x+\dfrac{1}{y}=6\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{y}=2\Rightarrow\dfrac{x}{y}=1\\x+\dfrac{1}{y}=-3\Rightarrow\dfrac{x}{y}=6\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+\dfrac{1}{y}=2\\\dfrac{x}{y}=1\end{matrix}\right.\) \(\Rightarrow...\)
29 tháng 11 2021
Cho hình bình hành ABCD,cạnh AB=a.AD=b .Tính AC^2+BD^2 theo a và b
giúp em với ạ
TM
11 tháng 4 2017
Bài 1)
Ta có:
A = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}\)
A < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
A < \(1-\dfrac{1}{8}\) = \(\dfrac{7}{8}\) < 1
Vậy A < 1
TM
12 tháng 4 2017
Bài 2)
Ta thấy:
\(\dfrac{2011}{2012+2013}< \dfrac{2011}{2012};\dfrac{2012}{2012+2013}< \dfrac{2012}{2013}\)
\(\Rightarrow\) \(\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}< \dfrac{2011}{2012}+\dfrac{2012}{2013}\)
\(\Rightarrow\) \(\dfrac{2011+2012}{2012+2013}< \dfrac{2011}{2012}+\dfrac{2012}{2013}\)
\(\Rightarrow\) A < B
Bài 3)
Ta có:
B = \(\left(1-\dfrac{1}{1}\right)\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)......\left(1-\dfrac{1}{20}\right)\)
= \(0.\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)......\left(1-\dfrac{1}{20}\right)\)
= 0
Bài 3)
Ta có:
A = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\)
\(\Rightarrow\) 2A = \(2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\right)\)
\(\Rightarrow\) 2A = \(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2011}}\)
\(\Rightarrow\) 2A - A = \(\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2011}}\right)\)-\(\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\right)\)
\(\Rightarrow\) A = 2 - \(\dfrac{1}{2^{2012}}\) = \(\dfrac{2^{2013}-1}{2^{2012}}\)
Bài 5)
\(\pi\) + 5 \(⋮\) \(\pi\) - 2
\(\Leftrightarrow\) \(\pi\) - 2 + 7 \(⋮\) \(\pi\) - 2
\(\Leftrightarrow\) 7 \(⋮\) \(\pi\) - 2 (vì \(\pi\) - 2 \(⋮\) \(\pi\) - 2)
\(\Leftrightarrow\) \(\pi\) - 2 \(\in\) Ư(7)
\(\Leftrightarrow\) \(\pi\) - 2 \(\in\) \(\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow\) \(\pi\) \(\in\) \(\left\{1;3;-5;9\right\}\)
8 tháng 3 2018
a) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow\)\(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+2^2+2^3\right)=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+4+8\right)=480\)
\(\Leftrightarrow\)\(2^x.15=480\)
\(\Rightarrow\)\(2^x=480:15\)
\(\Leftrightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy x = 5.
A
6 tháng 8 2017
\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\left(1+\dfrac{2012}{2}\right)+\left(1+\dfrac{2011}{3}\right)+...+\left(1+\dfrac{2}{2012}\right)+\left(1+\dfrac{1}{2013}\right)+1\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2012}+\dfrac{2014}{2013}+\dfrac{2014}{2014}\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)\)
\(\Leftrightarrow x=\dfrac{2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}}\)
\(\Leftrightarrow x=2014\)
Vậy \(x=2014\)
MV
6 tháng 8 2017
\(VP=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{1}{2013}\\ =\dfrac{2012}{2}+1+\dfrac{2011}{3}+1+...+\dfrac{1}{2013}+1+1\\ =\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2013}+\dfrac{2014}{2014}\\ =2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\)
\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\\ x=2014\)
Vậy x = 2014
29 tháng 6 2022
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\cdot x=\left(1+\dfrac{2011}{2}\right)+\left(1+\dfrac{2010}{3}\right)+...+\left(\dfrac{1}{2012}+1\right)+1\)
\(\Leftrightarrow x\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)=\dfrac{2013}{2}+\dfrac{2013}{3}+...+\dfrac{2013}{2013}\)
=>x=2013
1) \(x^2-6y^2=1\)
=> \(x^2-1=6y^2\)
=> \(y^2=\frac{x^2-1}{6}\)
Nhận thấy y^2 thuộc Ư của
\(\dfrac{x^2-1}{6}\)
=> \(y^2\) là số chẵn.
Mà y là số nguyên tố.
=> y = 2.
Thay vào:
=> \(x^2-1=\dfrac{4}{6}=24\)
=> \(x^2=25\)
=> \(x=5\)
Vậy: x = 5; y = 2.