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Nguyễn Vũ Lan Anh
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Nguyễn Vũ Lan Anh
3 tháng 5 2019 lúc 20:52

Giúp Mik ik mai nộp oy

SkillSkillSkill
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SkillSkillSkill
25 tháng 9 2018 lúc 11:17

Giúp mik với mn ơi

Kirigaya Kazuto
25 tháng 9 2018 lúc 12:02

1.\(\left(-\frac{6}{5}+\frac{6}{16}-\frac{6}{23}\right):\left(\frac{9}{5}-\frac{9}{16}+\frac{9}{23}\right)\)

\(=6\left(-\frac{1}{5}+\frac{1}{16}-\frac{1}{23}\right):\left(-9\right)\left(\frac{-1}{5}+\frac{1}{16}-\frac{1}{23}\right)\)

\(=6:\left(-9\right)=-\frac{2}{3}\)

2. \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{0.5-\frac{1}{3}+\frac{1}{4}}{-\frac{3}{2}+1-\frac{3}{4}}\)

\(=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{-3\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)

\(=\frac{3}{5}-\frac{1}{3}\)

\(=\frac{9}{13}-\frac{5}{15}=\frac{4}{15}\)

Lexiys
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Nobi Nobita
31 tháng 8 2020 lúc 8:52

b) \(\frac{12}{19}.\frac{7}{15}.\frac{-13}{17}.\frac{19}{12}.\frac{17}{13}=\frac{12}{19}.\frac{19}{12}.\frac{-13}{17}.\frac{17}{13}.\frac{7}{15}=1.\left(-1\right).\frac{7}{15}=\frac{-7}{15}\)

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★luffyッcute★(Team  ASL)
31 tháng 8 2020 lúc 9:03

\(-\frac{5}{7}.\frac{2}{11}+-\frac{5}{7}.\frac{9}{14}+\frac{12}{7}=-\frac{5}{7}.\left(\frac{2}{11}+\frac{9}{14}\right)+\frac{12}{7}=-\frac{5}{7}.\frac{127}{154}+\frac{12}{7}=-\frac{635}{1078}+\frac{12}{7}=\frac{1213}{1078}\)

\(\frac{12}{19}.\frac{7}{15}.-\frac{13}{17}.\frac{19}{12}.\frac{17}{13}=\left(\frac{12}{19}.\frac{19}{12}\right).\left(-\frac{13}{17}.\frac{17}{13}\right).\frac{7}{15}=1.-1.\frac{7}{15}=-\frac{7}{15}\)

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Nguyễn Thế Mãnh
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Nguyễn Huy Tú
14 tháng 9 2016 lúc 19:19

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) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)

Monkey D Luffy
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Nguyễn Hưng Phát
15 tháng 9 2016 lúc 12:39

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 \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)

Nguyễn Hưng Phát
15 tháng 9 2016 lúc 12:14

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:

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)

\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)

Choi Yuna
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Nguyễn Hữu Quang
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sakura
8 tháng 4 2017 lúc 7:31

ủng hộ mk nha mọi người

Lê Việt Hùng
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Mai Ngọc
15 tháng 3 2016 lúc 21:33

\(\left(-1\frac{1}{6}\right)\left(\frac{1-\frac{3}{5}+\frac{3}{11}-\frac{3}{13}}{\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}}\right)\left(\frac{4-\frac{4}{17}+\frac{4}{19}-\frac{4}{2013}}{5-\frac{5}{7}+\frac{5}{19}-\frac{5}{2013}}\right)\)

\(=-\frac{7}{6}.\left(\frac{3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}\right)}{\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}}\right):\left(\frac{4.\left(1-\frac{1}{7}+\frac{1}{19}-\frac{1}{2013}\right)}{5.\left(1-\frac{1}{7}+\frac{1}{19}-\frac{1}{2013}\right)}\right)\)

\(=-\frac{7}{6}.3:\frac{4}{5}=-\frac{7}{2}.\frac{5}{4}=-\frac{35}{8}\)

Nobita
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Nguyen tien Dung
8 tháng 4 2017 lúc 8:50

minh ko biet nua moi lop 4 thoi