TÍNH BIỂU THỨC
S = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{2016}-\frac{1}{2017}\)
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!
a,Tìm x biết
|3x-5|=|x+2|
b, tính giá trị biểu thức
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}\)
Nguyễn Tiến Đạt
a)\(|3x-5|=|x+2|\)
=> Ta có 2 trường hợp
*) TH1: 3x-5=x+2
=>3x-x=2+5
=>2x=7
=>x=7:2\(\Rightarrow x=\frac{7}{2}\)
*)TH2: -3x+5=x+2
\(\Rightarrow5-3x=x+2\)
\(\Rightarrow5-2=x+3x\)
\(\Rightarrow3=4x\)
\(\Rightarrow x=3:4\Rightarrow x=\frac{3}{4}\)
Vậy \(x\in\left\{\frac{7}{2};\frac{3}{4}\right\}\)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3+3\sqrt{4}}}+...+\frac{1}{2017\sqrt{2016}+2016\sqrt{2017}}\)
Tính giá trị của biểu thức .
\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)
Rút gọn biểu thức \(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2018}}\)
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{1+\left(1+\frac{2016}{2}\right)+\left(1+\frac{2015}{3}\right)+...+\left(1+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=2018\)
Ta có :
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\left(\frac{2017}{1}-1-1-...-1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=2018\)
Vậy \(A=2018\)
Chúc bạn học tốt ~
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2017}}\)
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 = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
tính giá trị biểu thức sau
B=(\(2017-\frac{1}{4}-\frac{2}{5}-\frac{3}{6}-\frac{4}{7}-...-\frac{2017}{2020}\)) : (\(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+\frac{1}{35}+...+\frac{1}{10100}\))
(ai nhanh, đúng, đủ mik tick lun)
Ta có
\(2017-\left(\frac{1}{4}+\frac{2}{5}+\frac{3}{6}+\frac{4}{7}+...+\frac{2017}{2020}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{4}+\frac{2}{5}+...+\frac{2017}{2020}\right)\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{2}{5}\right)+...+\left(1-\frac{2017}{2020}\right)\)
\(=\frac{3}{4}+\frac{3}{5}+....+\frac{3}{2020}\)
\(=\frac{3.5}{4.5}+\frac{3.5}{5.5}+\frac{3.5}{6.5}+...+\frac{3.5}{2020.5}\)
\(=3.5\left(\frac{1}{4.5}+\frac{1}{5.5}+\frac{1}{6.5}+...+\frac{1}{2020.5}\right)\)
\(=15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)\)
Thế vào ta có
\(\frac{15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)}{\frac{1}{20}+\frac{1}{25}+...+\frac{1}{10100}}=15\)
Được cập nhật 41 giây trước (17:23)
Ta có :
2017−(14 +25 +36 +47 +...+20172020 )
=(1+1+...+1)−(14 +25 +...+20172020 )
=(1−14 )+(1−25 )+...+(1−20172020 )
=34 +35 +....+32020
=3.54.5 +3.55.5 +3.56.5 +...+3.52020.5
=3.5(14.5 +15.5 +16.5 +...+12020.5 )
=15.(1
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
C = \(\frac{\left(a^{2016}+b^{2016}\right)^{2017}}{\left(c^{2016}+d^{2016}\right)^{2017}}\)= \(\frac{\left(a^{2017}-b^{2017}\right)^{2016}}{\left(c^{2017}-d^{2017}\right)^{2016}}\)
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
cs ng làm đung r
đag định lm
Tính nhanh : \(\frac{2017+\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}}\)