LP
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
DQ
0
14 tháng 7 2023
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
15 tháng 7 2023
Bài \(1\)
\(1)\) \(1-5+7-8+4-1+5-7+8\)
\(=(1-1)+(5-5)+(7-7)+(8-8)\)
\(=0+0+0+0\)
\(=0\)
\(2)\) \(14-23+(5+14)-(5-23)+17\)
\(=14-23+5+14-5+23+17\)
\(=(14+14)+(23-23)+(5-5)+17\)
\(=28+17\)
\(=45\)
\(3)\) \(12-44+9-3+14-19-9-12\)
\(=(12-12)+(9-9)+(14-44)+3\)
\(=-30+3\)
\(=-33\)
\(4)\) \(22-(4-8+12)+(-8-12+4)\)
\(=22-4+8-12-8-12+4\)
\(=22+(4+4)+(8-8)+(-12-12)\)
\(=22-24\)
\(=-2\)
27 tháng 7 2021
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
3 tháng 8 2015
a, \(\frac{-3}{7}+\frac{5}{13}-\frac{4}{7}+\frac{8}{13}\)
\(=\frac{-3}{7}-\frac{4}{7}+\frac{5}{13}+\frac{8}{13}\)
\(=-\frac{7}{7}+\frac{13}{13}=-1+1=0\)
b, \(\frac{-5}{14}-\frac{2}{-14}+\frac{1}{8}+\frac{1}{8}\)
\(=\frac{-5}{14}+\frac{2}{14}+\frac{1}{8}+\frac{1}{8}\)
\(=-\frac{3}{14}+\frac{1}{4}=\frac{1}{28}\)
c,\(-\frac{5}{13}-\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)\)
\(=-\frac{5}{13}-\frac{3}{13}-\frac{3}{5}+\frac{4}{10}\)
\(=-\frac{8}{13}-\frac{3}{5}+\frac{4}{10}=-\frac{79}{65}+\frac{4}{10}=-\frac{53}{65}\)
d, \(\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{4}{6}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)
\(=\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{2}{3}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)
\(=\left[\left(\frac{1}{8}-\frac{1}{2}-\frac{9}{7}-\frac{12}{7}+\frac{2}{3}\right)+\frac{5}{9}\right]\)
\(=-\frac{65}{24}+\frac{5}{9}=-2\frac{11}{72}\)
TL
B={[5^10. 7^3 - 25^2 . 49^2] : [(125 . 7)^3+5^9 . 14^3]} - {[2^12-4^6 . 9^2] : [(2^2.3)^6+8^4 . 3^5]
1
9 tháng 7 2023
Sửa đề: \(5^9\cdot49^2\)
\(=\dfrac{5^{10}\cdot7^3-5^9\cdot7^4}{5^9\cdot7^3+5^9\cdot14^3}-\dfrac{2^{12}-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}\)
\(=\dfrac{5^9\cdot7^3\left(5-7\right)}{5^9\cdot7^3\left(1+8\right)}-\dfrac{2^{12}\left(1-3^4\right)}{2^{12}\left(3^6+3^5\right)}=\dfrac{-2}{9}+\dfrac{80}{972}\)
=-34/243
SB
3
27 tháng 7 2023
\(\dfrac{9}{14}-\dfrac{5}{8}+\dfrac{3}{7}=\dfrac{36}{56}-\dfrac{35}{56}+\dfrac{24}{56}=\dfrac{25}{56}\)
\(\dfrac{2}{9}-\dfrac{7}{10}+\dfrac{4}{5}=\dfrac{20}{90}-\dfrac{63}{90}+\dfrac{72}{90}=\dfrac{29}{90}\)
\(\dfrac{8}{19}+\dfrac{7}{5}-\dfrac{9}{20}+\dfrac{11}{12}=\dfrac{8}{19}+\dfrac{84}{60}-\dfrac{27}{60}+\dfrac{55}{60}=\dfrac{8}{19}+\dfrac{112}{60}=\dfrac{8}{19}+\dfrac{28}{15}=\dfrac{652}{285}\)