K
3
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
24 tháng 1 2022
\(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,53+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}.\)
\(=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{53}{100}+\dfrac{1}{2}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+1-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}.\)
\(=\dfrac{\dfrac{263}{440}}{\dfrac{-1487}{1650}}+\dfrac{\dfrac{7}{4}}{\dfrac{35}{12}}=\dfrac{-3945}{5948}+\dfrac{3}{5}=\dfrac{-1881}{29740}.\)
22 tháng 10 2021
Bài 1:
a) \(\dfrac{19}{12}+\left|\dfrac{-5}{2}\right|+\left(\dfrac{3}{2}\right)^2=\dfrac{19}{12}+\dfrac{5}{2}+\dfrac{9}{4}\)
\(=\dfrac{19+5.6+9.3}{12}=\dfrac{76}{12}=\dfrac{19}{3}\)
b) \(\dfrac{2}{11}.\dfrac{16}{9}-\dfrac{2}{11}.\dfrac{7}{9}=\dfrac{2}{11}\left(\dfrac{16}{9}-\dfrac{7}{9}\right)=\dfrac{2}{11}.1=\dfrac{2}{11}\)
Bài 2:
Áp dụng t/c dtsbn:
\(\dfrac{a}{8}=\dfrac{b}{3}=\dfrac{a-b}{8-3}=\dfrac{55}{5}=11\)
\(\Rightarrow\left\{{}\begin{matrix}x=11.8=88\\b=11.3=33\end{matrix}\right.\)
ND
Nguyễn Đăng Nhân
CTVHS
18 tháng 9 2023
a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)
\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)
b)
\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)
\(=-14\)
c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)
\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)
17 tháng 9 2023
\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)
\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)
\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)
T
10 tháng 9 2023
\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)
\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)
\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)
\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)
\(=0,2-\dfrac{2}{3}\)
\(=-\dfrac{7}{15}\)
\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)
\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)
\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)
\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)
\(=\dfrac{13}{26}\)
\(=\dfrac{1}{2}\)
#\(Toru\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
10 tháng 9 2023
\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)
HK
2 tháng 11 2021
A= \(\dfrac{11}{9}\).\(\dfrac{-3}{2}\)- \(\dfrac{11}{9}\).\(\dfrac{15}{2}\)+(2021)0
= \(\dfrac{11}{9}\)(\(\dfrac{-3}{2}\)-\(\dfrac{15}{2}\)) + 1
= \(\dfrac{11}{9}\)(-9) + 1
= -11 + 1
= -10
2 tháng 11 2021
Lần sau bn ko đc phép chuyển \(\left(-2021\right)^0\) sang \(\left(2021\right)^0\) đou nha
12 tháng 12 2021
\(a,\dfrac{3}{5}+\dfrac{1}{5}.\dfrac{-17}{9}=\dfrac{3}{5}-\dfrac{17}{45}=\dfrac{27}{45}-\dfrac{17}{45}=\dfrac{10}{45}=\dfrac{2}{9}\\ b,\left(-\dfrac{4}{15}-\dfrac{18}{19}\right)-\left(\dfrac{20}{19}+\dfrac{11}{15}\right)=-\dfrac{4}{15}-\dfrac{18}{19}-\dfrac{20}{19}-\dfrac{11}{15}=\left(-\dfrac{4}{15}-\dfrac{11}{15}\right)-\left(\dfrac{18}{19}+\dfrac{20}{19}\right)=-1-2=-3\)
12 tháng 12 2021
\(a,=\dfrac{3}{5}+\left(-\dfrac{17}{45}\right)=\dfrac{2}{9}\)
\(b,=-\dfrac{4}{15}-\dfrac{18}{19}-\dfrac{20}{19}-\dfrac{11}{15}=-3\)
VV
25 tháng 10 2021
\(1\)/
\(a\)) \(=\left(\dfrac{7}{5}-\dfrac{8}{7}\right)+\dfrac{17}{5}:0,6\)
\(=\dfrac{9}{35}+\dfrac{17}{3}\)
\(=\dfrac{622}{105}\)
\(b\)) \(=\dfrac{11}{6}+\dfrac{-14}{15}\)
\(=\dfrac{9}{10}\)
\(c\)/ \(=\dfrac{7}{4}-\dfrac{2}{3}\)
\(=\dfrac{13}{12}\)
\(=\dfrac{-10}{55}+\dfrac{15}{55}=\dfrac{5}{55}=\dfrac{1}{11}\)
\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)