-6/5

\(=\dfrac{6}{5}\left(\dfrac{3}{13}-\dfrac{16}{13}\right)=-\dfrac{6}{5}\)

bằng âm 6/5

=7/11+5/6+4/11-1/6

=(7/11+4/11)+(5/6-1/6)

=1+2/3

=5/3

\(A,\dfrac{7}{11}+\dfrac{5}{6}-\dfrac{-4}{11}-\dfrac{1}{6}\)

\(= (\dfrac{7}{11}+\dfrac{-4}{11})-(\dfrac{5}{6}-\dfrac{1}{6})\)

\(=\dfrac{3}{11}-\dfrac{2}{3}\)

\(= \dfrac{9}{33}-\dfrac{22}{33}\)

\(= \dfrac{-13}{33}\)

\(B,\dfrac{5}{9}.\dfrac{7}{3}+\dfrac{5}{9}.\dfrac{9}{13}-\dfrac{5}{9}:\dfrac{13}{5}\)

\(=(\dfrac{5}{9}.\dfrac{5}{9}:\dfrac{5}{9}).\dfrac{7}{3}+\dfrac{9}{13}-\dfrac{13}{5}\)

\(=(\dfrac{5}{9}.\dfrac{7}{3})+(\dfrac{9}{13}-\dfrac{13}{5})\)

\(=(\dfrac{5}{9}.\dfrac{21}{9})+(\dfrac{45}{65}-\dfrac{169}{65})\)

\(=\dfrac{35}{27}+\dfrac{-124}{65}\)

\(=\dfrac{2275}{1755}+\dfrac{-3348}{1755}\)

\(=\dfrac{-1073}{1755}\)

Phương pháp giải:

Trừ nhẩm các số rồi điền kết quả thích hợp vào chỗ trống.

Lời giải chi tiết:

13 − 9 = 4            13 − 8 = 5

13 − 5 = 8            13 − 6 = 7

13 − 4 = 9            13 − 7 = 6

13 − 3 − 2 = 8      13 − 5 = 8

\(\dfrac{6}{13}.\dfrac{5}{7}+\dfrac{6}{13}.\dfrac{2}{7}+\dfrac{7}{13}=\dfrac{6}{13}.\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{7}{13}=\dfrac{6}{13}.1+\dfrac{7}{13}=\dfrac{6}{13}+\dfrac{7}{13}=\dfrac{13}{13}=1\)

#Monster

\(\dfrac{6}{13}\).(\(\dfrac{2}{7}+\dfrac{5}{7}\))+\(\dfrac{7}{13}\)

\(\dfrac{6}{13}\).1+\(\dfrac{7}{13}\)

\(\dfrac{6}{13}\)+\(\dfrac{7}{13}\)

1

mik làm ở bài dưới rồi nha.

a: \(\dfrac{5}{13}\left(\dfrac{6}{29}-\dfrac{26}{39}\right)-\dfrac{6}{29}\cdot\left(\dfrac{5}{13}-\dfrac{29}{6}\right)\)

\(=\dfrac{5}{13}\cdot\dfrac{6}{29}-\dfrac{5}{13}\cdot\dfrac{26}{39}-\dfrac{6}{13}\cdot\dfrac{5}{13}+\dfrac{6}{29}\cdot\dfrac{29}{6}\)

\(=\dfrac{-5}{39}\cdot2+1=1-\dfrac{10}{39}=\dfrac{29}{39}\)

b: \(\dfrac{1\cdot198+2\cdot197+3\cdot196+...+198\cdot1}{1\cdot2+2\cdot3+...+198\cdot199}\)

\(=\dfrac{1\left(199-1\right)+2\left(199-2\right)+...+198\cdot\left(199-198\right)}{1\left(1+1\right)+2\left(1+2\right)+...+198\left(1+198\right)}\)

\(=\dfrac{199\left(1+2+...+198\right)-\left(1^2+2^2+...+198^2\right)}{\left(1+2+...+198\right)+\left(1^2+2^2+...+198^2\right)}\)

\(=\dfrac{199\cdot\dfrac{198\cdot199}{2}-\dfrac{198\cdot\left(198+1\right)\cdot\left(2\cdot198+1\right)}{6}}{198\cdot\dfrac{199}{2}+\dfrac{198\left(198+1\right)\left(2\cdot198+1\right)}{6}}\)

\(=\dfrac{3\cdot198\cdot199^2-198\cdot199\cdot397}{6}:\dfrac{3\cdot198\cdot199+198\cdot199\cdot397}{6}\)

\(=\dfrac{198\cdot199\left(3\cdot199-397\right)}{198\cdot199\left(3+397\right)}\)

\(=\dfrac{200}{400}=\dfrac{1}{2}\)

pls mai mình phải nộp rồi

a: =-3/24-40/24

=-43/24

b: \(=\dfrac{6}{54}\cdot\dfrac{49}{35}=\dfrac{1}{9}\cdot\dfrac{7}{5}=\dfrac{7}{45}\)

c: \(=\dfrac{6}{5}+\dfrac{4}{3}=\dfrac{18+20}{15}=\dfrac{38}{15}\)

d: \(=\dfrac{31}{17}-\dfrac{14}{17}-\dfrac{5}{13}-\dfrac{8}{13}=1-1=0\)

\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)

\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)

\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)

\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)

\(=-\dfrac{38}{42}\)

\(=-\dfrac{19}{21}\)

\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)

\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)

\(=\dfrac{13}{13}+11\)

\(=1+11\)

\(=12\)

\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)

\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)

\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)

\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)

\(=\dfrac{28}{7}-\dfrac{31}{9}\)

\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)

\(=\dfrac{252}{63}-\dfrac{217}{63}\)

\(=\dfrac{35}{63}\)

\(=\dfrac{5}{9}\)

\(5,\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\left(\dfrac{2}{3}-\dfrac{3}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)

\(=\dfrac{-15}{24}+\dfrac{1}{2}\)

\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)

\(=\dfrac{-15}{24}+\dfrac{12}{24}\)

\(=\dfrac{-3}{24}\)

\(=-\dfrac{1}{8}\)

\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)

\(=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}\)

\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)

\(=-1+1-\dfrac{3}{7}\)

\(=-\dfrac{3}{7}\)

\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)

\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)

\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)

\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)

\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)

\(=\dfrac{6}{5}\times\dfrac{20}{7}\)

\(=\dfrac{120}{35}\)

\(=\dfrac{24}{7}\)

\(\dfrac{4}{5}\)+\(\dfrac{-5}{4}\)=\(\dfrac{0}{4}\)

\(\dfrac{-1}{3}\)+\(\dfrac{2}{5}-\dfrac{5}{6}\)=\(\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}\)=\(\dfrac{-23}{30}\)

\(\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}\)=\(\dfrac{2}{3}-\dfrac{2}{5}\)=\(\dfrac{10}{15}-\dfrac{6}{15}\)=\(\dfrac{4}{15}\)

a: \(=\dfrac{16-25}{20}=\dfrac{-9}{20}\)

b: \(=\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}=-\dfrac{23}{30}\)

c: \(=\dfrac{2}{3}-\dfrac{1}{5}\cdot2=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10-6}{15}=\dfrac{4}{15}\)

d: \(=\dfrac{12-25}{30}\cdot\dfrac{15}{8}-1=\dfrac{-13}{16}-1=\dfrac{-29}{16}\)

e: \(=\dfrac{-3}{13}-\dfrac{10}{13}-\dfrac{7}{17}+\dfrac{24}{17}-\dfrac{5}{19}-\dfrac{14}{19}=-1\)