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Khách
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TQ
2
11 tháng 3 2020
x=(1+2+3-4-5-6)+...+(97+98+99-100-101-102)
x=-9+...+-9
x=-9.17
x=-153
9 tháng 3 2018
các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 2 2024
a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)
-\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)
-6 < \(x\) < -4
vì \(x\) \(\in\) Z nên \(x\) = -5
NN
1 tháng 8 2020
a) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+..........+\frac{2}{73.75}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.......+\frac{1}{73}-\frac{1}{75}\)
\(=\frac{1}{3}-\frac{1}{75}=\frac{8}{25}\)
c) \(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+..........+\frac{4}{64.66}\)
\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+..........+\frac{2}{64.66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+.....+\frac{1}{64}-\frac{1}{66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{66}\right)=2.\frac{31}{132}=\frac{31}{66}\)
NN
1 tháng 8 2020
d) \(\frac{9}{5.8}+\frac{9}{8.11}+\frac{9}{11.14}+........+\frac{9}{497.500}\)
\(=3.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+..........+\frac{3}{497.500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+......+\frac{1}{497}-\frac{1}{500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{500}\right)=3.\frac{99}{500}=\frac{297}{500}\)
e) \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+......+\frac{1}{93.95}\)
\(=\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+........+\frac{2}{93.95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+........+\frac{1}{93}-\frac{1}{95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{95}\right)=\frac{1}{2}.\frac{18}{95}=\frac{9}{95}\)
g) \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..........+\frac{1}{200.203}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{200.203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{200}-\frac{1}{203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{203}\right)=\frac{1}{3}.\frac{201}{406}=\frac{67}{406}\)
15 tháng 8 2018
(1) 5/2 x (3/4 - 1/3) -11/12+1/4 = 5/2 x (9/12-4/12) -11/12 +3/12=5/2 x 5/12 -11/12 +3/12 = 25/24 - 11/12 +3/12=25/24 -22/24 + 6/24=9/12=3/4
(2) 11/12 : (4/10 + 3/5) + (5/6 - 1/2) x2/3
= 11/12 : (4/10 + 6/10) + (5/6-3/6) x 2/3
= 11/12:1 +1/2 x 2/3
= 11/12 + 2/3 =11/12+8/12 = 19/12
2 tháng 4 2017
1) \(\frac{5}{8}.\frac{7}{3}-\frac{5}{2}.\frac{1}{8}=\frac{5}{8}.\frac{7}{3}-\frac{5}{8}.\frac{1}{2}=\frac{5}{8}\left(\frac{7}{3}-\frac{1}{2}\right)=\frac{5}{8}.\frac{11}{6}=\frac{55}{48}\)
2) \(\frac{21}{10}.\frac{3}{4}-\frac{21}{10}.\frac{3}{4}=\frac{21}{10}\left(\frac{3}{4}-\frac{3}{4}\right)=\frac{21}{10}.0=0\)
3) \(\frac{-4}{11}:\frac{-6}{11}=\frac{-4}{11}.\frac{-11}{6}=\frac{-4.\left(-11\right)}{11.6}=\frac{-4.\left(-1\right)}{1.6}=\frac{4}{6}=\frac{2}{3}\)
4)\(\frac{2}{7}.\frac{14}{3}-1=\frac{2.14}{7.3}-1=\frac{2.2}{1.3}-1=\frac{4}{3}-1=\frac{1}{3}\)
5)\(\frac{4}{7}:\left(\frac{1}{5}.\frac{4}{7}\right)=\frac{4}{7}:\frac{4}{35}=\frac{4}{7}.\frac{35}{4}=\frac{4.35}{7.4}=\frac{1.5}{1.1}=5\)
6) \(\frac{12}{7}.\frac{7}{4}+\frac{35}{11}:\frac{245}{121}=\frac{12.7}{7.4}+\frac{35}{11}.\frac{121}{245}=\frac{3.1}{1.1}+\frac{35}{11}.\frac{121}{245}=3+\frac{35}{11}.\frac{121}{245}=3+\frac{35.121}{11.245}=\frac{1.11}{1.7}=\frac{11}{7}\)
5.36536796537
cách làm như thế nào