\(\dfrac{9}{3}+\dfrac{3}{3}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
3 tháng 3 2023
\(\dfrac{24\cdot47-23}{24+47\cdot23}\cdot\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\\ =\dfrac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\dfrac{3\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}{9\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}\\ =\dfrac{24\cdot24+24\cdot23-23\cdot1}{24+24\cdot23+23\cdot23}\cdot\dfrac{1}{3}\\ =\dfrac{23\left(24-1\right)+24\cdot24}{24\left(1+23\right)+23\cdot23}\cdot\dfrac{1}{3}=\dfrac{23\cdot23+24\cdot24}{24\cdot24+23\cdot23}\cdot\dfrac{1}{3}\\ =1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)
27 tháng 5 2022
\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)
=1/64
HT
Hồ Tuyết Anh
VIP
16 tháng 3 2022
\(e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
\(f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)
\(=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}\)
\(=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}\)
\(g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)\)
\(=\dfrac{5}{9}\cdot1=\dfrac{5}{9}\)
8 tháng 3 2023
1, \(\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}+1\right)\)
\(=\dfrac{3}{4}.\dfrac{6-1+15}{15}=\dfrac{3}{4}.\dfrac{20}{15}=\dfrac{3}{4}.\dfrac{4}{3}=1\)
2, \(\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{3}.\dfrac{40}{9}=\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{9}.\dfrac{40}{3}\)
\(=\dfrac{4}{9}.\left[\left(-\dfrac{13}{3}\right)+\dfrac{40}{3}\right]=\dfrac{4}{9}.9=4\)
3, \(\dfrac{4}{9}-\dfrac{2}{3}.\left(\dfrac{4}{5}+\dfrac{1}{2}\right)=\dfrac{2}{3}\left(\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{3}.\dfrac{20-24-15}{30}=\dfrac{2}{3}.\left(-\dfrac{19}{30}\right)=-\dfrac{19}{45}\)
TH
Thầy Hùng Olm
Manager
VIP
8 tháng 3 2023
1. \(\dfrac{3}{4}.\left(\dfrac{6}{15}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\dfrac{1}{3}+\dfrac{3}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1\)
21 tháng 2 2022
\(=\dfrac{85}{18}:\dfrac{85}{9}-\dfrac{136}{45}:\dfrac{136}{15}=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\)
74
22 tháng 4
Để tính tổng của biểu thức này, chúng ta cần thực hiện các phép cộng và trừ theo thứ tự từ trái sang phải.
\[4 + \frac{5}{6} - \frac{1}{9} \times \frac{1}{10} - \frac{7}{12} + \frac{1}{36} - 3 - \frac{1}{5} + \frac{1}{3} - \frac{1}{9} \times \frac{9}{5} + 1 - \frac{1}{3}\]
Đầu tiên, chúng ta sẽ làm các phép tính liên quan đến phân số:
\[= 4 + \frac{5}{6} - \frac{1}{90} - \frac{7}{12} + \frac{1}{36} - 3 - \frac{1}{5} + \frac{1}{3} - \frac{1}{5} + 1 - \frac{1}{3}\]
Tiếp theo, chúng ta sẽ tổng hợp các phân số:
\[= 4 + \frac{5}{6} - \frac{1}{90} - \frac{35}{90} + \frac{5}{180} - 3 - \frac{18}{90} + \frac{60}{180} - \frac{18}{90} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{1}{90} - \frac{35}{90} + \frac{5}{180} - 3 - \frac{2}{10} + \frac{10}{30} - \frac{2}{10} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{1}{90} - \frac{35}{90} + \frac{5}{180} - 3 - \frac{1}{5} + \frac{1}{3} - \frac{1}{5} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{36 + 35}{90} + \frac{5}{180} - 3 - \frac{1}{5} + \frac{2}{6} - \frac{1}{5} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - 3 - \frac{1}{5} + \frac{1}{3} - \frac{1}{5} + 1 - \frac{1}{3}\]
Tiếp theo, chúng ta sẽ tính tổng các số nguyên:
\[= 4 - 3 + 1\]
Cuối cùng, chúng ta sẽ tổng hợp các phân số:
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{1}{5} + \frac{1}{3} - \frac{1}{5} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{30}{90}\]
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]
\[= 4 + \frac{5}{6}
27 tháng 2 2022
\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)
\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)
\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)
\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)
\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)
12,3
\(\dfrac{9}{3}+\dfrac{3}{3}=3+1=4\)