Cho S = \(\frac{2}{\frac{1}{2016}+\frac{3}{2017}+\frac{5}{2018}+...+\frac{47}{2039}}\)
Chứng minh 7 < S < 8
Cho S = \(\frac{2}{\frac{1}{2016}+\frac{2}{2017}+\frac{3}{2018}+...+\frac{47}{2039}}\)
Chứng minh 7 < S < 8
\(A=\frac{1}{2017}+\frac{2}{2017^2}+\frac{3}{2017^3}+...+\frac{2017}{2017^{2017}}+\frac{2018}{2017^{2018}}\). Chứng minh rằng : A < \(\frac{2017}{2016^2}\)
a)\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
b)\(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\)
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy : \(x=-2020\)
Chúc bạn học tốt !!
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)
Vậy x = -2020
b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)
Vậy x = -2010
Cho S = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
P = \(1+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}\)
Chứng minh rằng: \(\left(S-P\right)^{2018}=1\)
\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-2.\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=P-1\)
\(\Rightarrow\left(S-P\right)^{2018}=\left(P-1-P\right)^{2018}=\left(-1\right)^{2018}=1\)
Câu 1. Tính hợp lý giá trị các biểu thức sau :
a. A = ( 689 - 31 ) - ( 269 - 131 )
b. B = \(\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}+1\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}\right)-\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}+1\right)\)c. C = \(1-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
cho S=\(\frac{1}{3}\)-\(\frac{2}{3^2}\)+\(\frac{3}{3^3}\)-\(\frac{4}{3^4}\)...+\(\frac{2017}{3^{2017}}\)-\(\frac{2018}{3^{2018}}\).chứng minh S<\(\frac{3}{16}\)
A=\(\frac{\frac{1}{2018}+\frac{2}{2017}+\frac{3}{2016}+....+\frac{2017}{2}+\frac{2018}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2019}}\). Chứng minh rằng A là số nguyên
Mong mọi người giúp
Tim x
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
2/ tim x
\(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7} +\frac{x+2018}{8}\)
3/ tim x
\(\frac{1}{3}+\frac{1}{6}+\frac{99}{101}+\frac{1}{15}+... +\frac{1}{x\left(2x+1\right)}=\frac{1}{10}\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
Cảm ơn bạn rất nhiều mình đã hiểu rồi
Chúc bạn học tốt nhé
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2017}}\)
chứng tỏ A<1
2,
\(S=2+2^2+2^3+...+2^{99}\)
C/t: S chia hết cho 7, 31
3,
\(A=1+5+5^2+5^3+5^4+5^5+...+5^{99}+5^{100}\)
Tính A
4,
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)<1
5,
CHỨNG tỏ rằng các p/s tối giản vs mọi số tự nhiên n
a,\(\frac{n+1}{2n+3}\)b,\(\frac{2n+3}{4n+8}\)
6,
a,TÍnh A và B
A=\(\frac{2016^{2016}+2}{2016^{1016}-1}\)B=\(\frac{2016^{2016}}{2016^{2016}-3}\)
b, tính
C=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)
LÀm NHANH Hộ MK ,MAi mk Phải Nộp.
2/
S = 2 + 22 + 23 +...+ 299
= (2+22+23) +...+ (297+298+299)
= 2(1+2+22)+...+297(1+2+22)
= 2.7 +...+ 297.7
= 7(2+...+297) chia hết cho 7
S = 2+22+23+...+299
= (2+22+23+24+25)+...+(295+296+297+298+299)
= 2(1+2+22+23+24)+...+295(1+2+22+23+24)
= 2.31+...+295.31
= 31(2+...+295) chia hết cho 31
3/
A = 1+5+52+....+5100 (1)
5A = 5+52+53+...+5101 (2)
Lấy (2) - (1) ta được
4A = 5101 - 1
A = \(\frac{5^{101}-1}{4}\)
4/
Đặt A là tên của biểu thức trên
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
........
\(\frac{1}{8^2}< \frac{1}{7.8}=\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}< 1\)
Vậy...
5/
a, Gọi UCLN(n+1,2n+3) = d
Ta có : n+1 chia hết cho d => 2(n+1) chia hết cho d => 2n+2 chia hết cho d
2n+3 chia hết cho d
=> 2n+2 - (2n+3) chia hết cho d
=> -1 chia hết cho d => d = {-1;1}
Vậy...
b, Gọi UCLN(2n+3,4n+8) = d
Ta có: 2n+3 chia hết cho d => 2(2n+3) chia hết cho d => 4n+6 chia hết cho d
4n+8 chia hết cho d
=> 4n+6 - (4n+8) chia hết cho d
=> -2 chia hết cho d => d = {1;-1;2;-2}
Mà 2n+3 lẻ => d lẻ => d khác 2;-2 => d = {1;-1}
Vậy...
6/
a,Vì B > 1
\(\Rightarrow B=\frac{2016^{2016}}{2016^{2016}-3}>\frac{2016^{2016}+2}{2016^{2016}-3+2}=\frac{2016^{2016}+2}{2016^{2016}-1}=A\)
Vậy A < B
b, C = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)