Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
1: =1-1+5-5+7-7+8-8=0
2: =14-23+5+14-5+23+17
=28+17=45
3: =12-12+9-9+14-44-3=-33
4: =22-8-8-12+4
=22-16-8
=-2
Bài \(1\)
\(1)\) \(1-5+7-8+4-1+5-7+8\)
\(=(1-1)+(5-5)+(7-7)+(8-8)\)
\(=0+0+0+0\)
\(=0\)
\(2)\) \(14-23+(5+14)-(5-23)+17\)
\(=14-23+5+14-5+23+17\)
\(=(14+14)+(23-23)+(5-5)+17\)
\(=28+17\)
\(=45\)
\(3)\) \(12-44+9-3+14-19-9-12\)
\(=(12-12)+(9-9)+(14-44)+3\)
\(=-30+3\)
\(=-33\)
\(4)\) \(22-(4-8+12)+(-8-12+4)\)
\(=22-4+8-12-8-12+4\)
\(=22+(4+4)+(8-8)+(-12-12)\)
\(=22-24\)
\(=-2\)
*\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left[\frac{1}{14}+\frac{1}{7}-\left(-\frac{3}{35}\right)\right].\frac{4}{3}}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right).\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right).\frac{4}{3}}=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{4}{3}}=\frac{\frac{-1}{12}}{\frac{14}{35}}=-\frac{1}{12}.\frac{35}{14}=\frac{-35}{168}\)
*\(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{63}{10}.12-21.\frac{18}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{378}{5}-\frac{378}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)
1) Ta có: \(\frac{-4}{7}-\frac{11}{19}+\frac{13}{19}\cdot\frac{-3}{7}+\frac{2}{19}:\frac{-7}{4}\)
\(=\frac{-4}{7}-\frac{11}{19}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-76}{133}-\frac{77}{133}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-200}{133}\)
2) Ta có: \(\left(\frac{-4}{9}+\frac{3}{5}\right):\frac{1}{\frac{1}{5}}+\left(\frac{1}{5}-\frac{5}{9}\right):\frac{1}{\frac{1}{5}}\)
\(=\left(\frac{-4}{9}+\frac{3}{5}\right)\cdot\frac{1}{5}+\left(\frac{1}{5}-\frac{5}{9}\right)\cdot\frac{1}{5}\)
\(=\frac{1}{5}\left(\frac{-4}{9}+\frac{3}{5}+\frac{1}{5}-\frac{5}{9}\right)\)
\(=\frac{1}{5}\left(-1+\frac{4}{5}\right)\)
\(=\frac{1}{5}\cdot\frac{-1}{5}=\frac{-1}{25}\)
3) Ta có: \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\)
\(=\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\)
\(=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\)
\(=\frac{27}{70}\)
4) Ta có: \(\frac{2}{7}-\left(-\frac{13}{15}+\frac{4}{9}\right)-\left(\frac{5}{9}-\frac{2}{15}\right)\)
\(=\frac{2}{7}+\frac{13}{15}-\frac{4}{9}-\frac{5}{9}+\frac{2}{15}\)
\(=\frac{2}{7}+1-1=\frac{2}{7}\)
