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28 tháng 6 2019
a) Ta có:
C = 5/18 + 8/19 - 7/21 + (-10/36 + 11/19 + 1/3) - 5/8
C = 5/18 + 8/19 - 1/3 - 5/18 + 11/19 + 1/3 - 5/8
C = (5/18 - 5/18) + (8/19 + 11/19) - (1/3 - 1/3) - 5/8
C = 1 - 5/8
c = 3/8
b) F = 15/14 - (17/23 - 80/87 + 5/4) + (17/23 - 15/14 + 1/4)
F = 15/14 - 17/23 + 80/87 - 5/4 + 17/23 - 15/14 + 1/4
F = (15/14 - 15/14) - (17/23 - 17/23) + 80/87 - (5/4 - 1/4)
F = 80/87 - 1
F = -7/87
c) G = 1/25 - 4/27 + (-23/27 + -1/25 - 5/43) + 5/43 - 4/7
G = 1/25 - 4/27 - 23/27 - 1/25 - 5/43 + 5/43 - 4/7
G = (1/25 - 1/25) - (4/27 + 23/27) - (5/43 - 5/43) - 4/7
G = -1 - 4/7 = -11/7
US
2
18 tháng 10 2018
=22/17 +16/27 -5/17 +0,5+7/23
=1+0,5+16/27+7/23
=2977/1242
18 tháng 10 2018
\(1\frac{5}{17}+\frac{16}{27}-\frac{5}{17}+0.5+\frac{7}{23}\)
\(=1+\frac{5}{17}+\frac{16}{27}-\frac{5}{17}+\frac{1}{2}+\frac{7}{23}\)
\(=1+\frac{16}{27}+\frac{1}{2}+\frac{7}{23}+\left(\frac{5}{17}-\frac{5}{17}\right)\)
\(=\frac{43}{27}+\frac{37}{46}+0\)
\(=\frac{1978}{1242}+\frac{999}{1242}\)
\(=\frac{2977}{1242}\)
14 tháng 9 2016
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)
15 tháng 9 2016
Làm tiếp:
\(=\left(1+\frac{1}{2}+.....+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+.........+\frac{1}{2017}\)
\(\Rightarrow\frac{\frac{1}{1009}+....+\frac{1}{2017}}{1-\frac{1}{2}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}=1\)
Bài 2:
Đặt \(A=\frac{1}{2^2}+.......+\frac{1}{2^{800}}\)
\(4A=1+\frac{1}{2^2}+.....+\frac{1}{2^{798}}\)
\(\Rightarrow4A-A=1-\frac{1}{2^{800}}\)
\(\Rightarrow3A=1-\frac{1}{2^{800}}< 1\Rightarrow A< \frac{1}{3}\)
Vậy \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)
15 tháng 9 2016
Bài 1:Tính
a, Xét biểu thức \(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).........\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)..........\left(1+\frac{n+2}{n}\right)}\) với\(n\in N\)
Ta có:\(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).......\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)......\left(1+\frac{n+2}{n}\right)}\)
\(=\frac{\frac{n+1}{1}.\frac{n+2}{2}........\frac{2n+2}{n+2}}{\frac{n+3}{1}.\frac{n+4}{2}.........\frac{2n+2}{n}}\)
\(=\frac{\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right)}{1.2.3.........\left(n+2\right)}}{\frac{\left(n+3\right)\left(n+4\right)........\left(2n+2\right)}{1.2.3.........n}}\)
\(=\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right).1.2.3.......n}{\left(n+3\right)\left(n+4\right)........\left(2n+2\right).1.2.3......\left(n+2\right)}\)
\(=\frac{\left(n+1\right)\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}=1\)
Áp dụng vào bài toán ta có đáp số là:1
b, \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=-\frac{2}{3}\)
c,\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{\frac{1}{3}}{\frac{1}{4}}=12\)
d,\(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}=\frac{2}{13}\)
e,Xét mẫu số ta có:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)
30 tháng 9 2021
A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 8 2024
C = 2-3 + (52)3.5-3 + 4-3.16 - 2.32 - 105.(\(\dfrac{24}{51}\))0
C = \(\dfrac{1}{8}\) + 56.5-3 + 4-3.42 - 2.9 - 105.1
C = \(\dfrac{1}{8}\) + 53 + \(\dfrac{1}{4}\) - 18 - 105
C = (\(\dfrac{1}{8}\) + \(\dfrac{1}{4}\)) - (105 - 125 + 18)
C = \(\dfrac{3}{8}\) - (-20 + 18)
C = \(\dfrac{3}{8}\) + 2
C = \(\dfrac{19}{8}\)
NQ
3
VM
17 tháng 1 2020
b) \(\frac{\frac{2}{3}+\frac{5}{7}+\frac{4}{21}}{\frac{5}{6}+\frac{11}{7}-\frac{7}{21}}\)
\(=\frac{\frac{29}{21}+\frac{4}{21}}{\frac{101}{42}-\frac{7}{21}}\)
\(=\frac{\frac{11}{7}}{\frac{29}{14}}\)
\(=\frac{22}{29}.\)
Chúc bạn học tốt!
23 tháng 7 2016
\(F=\frac{15}{14}-\left(\frac{17}{23}-\frac{80}{87}+\frac{5}{4}\right)+\left(\frac{17}{23}-\frac{15}{4}+\frac{1}{4}\right)\)
\(F=\frac{15}{14}-\frac{17}{23}+\frac{80}{87}-\frac{5}{4}+\frac{17}{23}-\frac{15}{4}+\frac{1}{4}\)
\(F=\frac{15}{14}+\left(\frac{-17}{23}+\frac{17}{23}\right)+\left(\frac{-5}{4}+\frac{-15}{4}+\frac{1}{4}\right)+\frac{80}{87}\)
\(F=\frac{15}{14}+0+\frac{-19}{4}+\frac{80}{87}\)
23 tháng 7 2016
\(\frac{15}{14}-\left(\frac{17}{23}-\frac{80}{87}+\frac{5}{4}\right)+\left(\frac{17}{23}-\frac{15}{4}+\frac{1}{4}\right)\)
\(=\frac{15}{14}-\frac{8561}{8004}+\left(-\frac{127}{46}\right)\)
\(=-2,759031199\)
-5/17+5/23+7/17+-1/23+-2/17+19/23
=(-5/17+7/17+-2/17)+(5/23+-1/23+19/23)
=0+1
=1
\(=1\)
HOK TỐT