= 811123,1263612567352132183321213 đó

rảnh đi ngồi hành hạ cái bàn phím, ghê ta

chế nào thấy thương cái bộ bàn phím thì đi qua nhớ em nhá

Ta tính nhẩm như sau:

3x4=12     2x5=10    5x3=15     4x2=8

12:3=4      10:2=5    15:3=5      8:2=4

12:4=3      10:5=2     15:5=3     8:4=2

3x4 = 12               2x5 = 10        5x3 = 15                  4x2 =8

12 : 3 = 4              10 : 2 = 5       15 : 3 = 5                8 : 2 = 4

12 : 4 =3               10 :5 = 2        15 : 5 = 3                8 : 4 = 2

   \(\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{2}{4}\)

= \(\dfrac{24}{60}\) + \(\dfrac{40}{60}\) + \(\dfrac{30}{60}\)

= \(\dfrac{64}{60}\) + \(\dfrac{30}{60}\)

= \(\dfrac{47}{30}\)

\(\dfrac{2}{6}+\dfrac{3}{12}\)

= \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\)

= \(\dfrac{7}{12}\)

\(\dfrac{5}{6}\) + \(\dfrac{1}{3}\)

= \(\dfrac{5}{6}\) + \(\dfrac{2}{6}\)

= \(\dfrac{7}{6}\)

   \(\dfrac{1}{3}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{6}\)

= \(\dfrac{4}{12}\) + \(\dfrac{5}{12}\) + \(\dfrac{10}{12}\)

= \(\dfrac{9}{12}\) + \(\dfrac{10}{12}\)

= \(\dfrac{19}{12}\)

    \(\dfrac{5}{8}\) + \(\dfrac{4}{7}\)

= \(\dfrac{35}{56}\) + \(\dfrac{32}{56}\)

=  \(\dfrac{67}{56}\) 

\(\dfrac{7}{3}\) + \(\dfrac{8}{7}\)

= \(\dfrac{49}{21}\) + \(\dfrac{24}{21}\)

= \(\dfrac{73}{21}\)

   \(\dfrac{1}{5}+\dfrac{5}{35}\)

= \(\dfrac{7}{35}\) + \(\dfrac{5}{35}\)

= \(\dfrac{12}{35}\) 

cách giải bài này như nào: 5/20+1/4+6/24+8/32=bao nhiêu

    \(\dfrac{4}{5}\) : (\(\dfrac{4}{5}\) .- \(\dfrac{5}{4}\)) : (\(\dfrac{16}{25}\) - \(\dfrac{1}{5}\))

=  \(\dfrac{4}{5}\) : (-1) : (\(\dfrac{16}{25}\) - \(\dfrac{5}{25}\))

= -\(\dfrac{4}{5}\)  : \(\dfrac{11}{25}\)

= - \(\dfrac{4}{5}\) x \(\dfrac{25}{11}\)

=  - \(\dfrac{20}{11}\)

    \(\dfrac{11}{12}\) : (\(\dfrac{7}{9}\) + \(\dfrac{-1}{3}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{5}{15}\))

= \(\dfrac{11}{12}\) : (\(\dfrac{7}{9}\) - \(\dfrac{3}{9}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{1}{3}\))

= \(\dfrac{11}{12}\) : \(\dfrac{4}{9}\) - \(\dfrac{1}{3}\)

= \(\dfrac{11}{12}\) x \(\dfrac{9}{4}\) - \(\dfrac{1}{3}\)

= \(\dfrac{99}{48}\) - \(\dfrac{1}{3}\)

= \(\dfrac{99}{48}\) - \(\dfrac{16}{48}\)

= \(\dfrac{83}{48}\)  

\(\dfrac{4}{5}\): (\(\dfrac{4}{5}\).-\(\dfrac{5}{4}\)) : (\(\dfrac{16}{25}\) - \(\dfrac{1}{5}\))

=\(\dfrac{4}{5}\) x - 1: (\(\dfrac{16}{25}\) - \(\dfrac{5}{25}\))

= - \(\dfrac{4}{5}\) : \(\dfrac{11}{25}\)

= - \(\dfrac{4}{5}\) x \(\dfrac{25}{11}\)

= - \(\dfrac{20}{11}\) 

\(\dfrac{11}{12}\): (\(\dfrac{7}{9}\) + - \(\dfrac{1}{3}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{5}{15}\))

= \(\dfrac{11}{12}\) : (\(\dfrac{7}{9}\) - \(\dfrac{3}{9}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{1}{3}\))

= \(\dfrac{11}{12}\) : \(\dfrac{4}{9}\) - \(\dfrac{1}{3}\)

= \(\dfrac{11}{12}\) x \(\dfrac{9}{4}\) - \(\dfrac{1}{3}\)

= \(\dfrac{99}{48}\) - \(\dfrac{16}{48}\)

= \(\dfrac{83}{48}\) 

- \(\dfrac{2}{5}\).(\(\dfrac{9}{8}\) + \(\dfrac{16}{32}\) - \(\dfrac{3}{4}\)) - \(\dfrac{9}{10}\)

= - \(\dfrac{2}{5}\).(\(\dfrac{9}{8}\) + \(\dfrac{4}{8}\) - \(\dfrac{6}{8}\)) - \(\dfrac{9}{10}\)

= - \(\dfrac{2}{5}\).(\(\dfrac{13}{8}\) - \(\dfrac{6}{8}\)) - \(\dfrac{9}{10}\)

= - \(\dfrac{2}{5}\).\(\dfrac{7}{8}\) - \(\dfrac{9}{10}\)

= - \(\dfrac{7}{20}\) - \(\dfrac{18}{20}\)

= - \(\dfrac{5}{4}\)

Ko bết

Phương pháp giải:

Nhẩm lại bảng nhân và chia trong phạm vi đã học rồi điền kết quả vào chỗ trống.

Lời giải chi tiết:

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.4:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.4}=-6\)

= -4 đúng không

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.\dfrac{4}{3}:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.3.4}=\dfrac{-6}{3}=-2\)

Vậy...

\(B=8+\left(\dfrac{1}{5}\cdot5\right)^4=8+1=9\)