1:

a: =8/7-5/88=669/616

b: \(=1+\dfrac{2}{9}\cdot\dfrac{3}{7}-\dfrac{10}{7}=1+\dfrac{2}{21}-\dfrac{10}{7}\)

\(=\dfrac{21+2-30}{21}=\dfrac{-7}{21}=\dfrac{-1}{3}\)

c: \(=\dfrac{11}{3}-\dfrac{6}{7}+4=\dfrac{77-18+84}{21}=\dfrac{143}{21}\)

Bài 2:

a: =>9/4-x=5/11*2=10/11

=>x=9/4-10/11=59/44

b: =>2/9:x=19/21

=>x=2/9:19/21=14/57

1: =72/90+65/90=137/90

2: =24/56-77/56=-53/56

3: =-7/10+4/5=1/10

4: =15/100-4/100=11/100

5: =4/6-5/6=-1/6

6: =10/40-15/40-76/40=-81/40

7: =-9/10+7/18

=-81/90+35/90=-46/90=-23/45

8: =27/90-55/90=-28/90=-14/45

9: =36/60-50/60-35/60=-49/60

10: =-4/9+5/6-3/8

=-32/72+60/72-27/72

=1/72

\(1,\dfrac{4}{5}+\dfrac{13}{18}=\dfrac{72}{90}+\dfrac{65}{90}=\dfrac{137}{90}\)

\(2,\dfrac{3}{7}-\dfrac{11}{8}=\dfrac{24}{56}-\dfrac{77}{56}=\dfrac{-53}{56}\)

\(3,-\dfrac{7}{10}-\left(-\dfrac{4}{5}\right)=-\dfrac{7}{10}-\left(-\dfrac{8}{10}\right)=\dfrac{1}{10}\)

\(4,\dfrac{3}{20}-\dfrac{1}{25}=\dfrac{75}{500}-\dfrac{20}{500}=\dfrac{55}{500}=\dfrac{11}{100}\)

\(5,\dfrac{2}{3}-\dfrac{5}{6}=\dfrac{12}{18}-\dfrac{15}{18}=-\dfrac{3}{18}=-\dfrac{1}{6}\)

\(6,\dfrac{1}{4}+\left(-\dfrac{3}{8}\right)-\dfrac{19}{10}=\dfrac{8}{32}+\left(-\dfrac{12}{32}\right)-\dfrac{19}{10}=-\dfrac{4}{32}-\dfrac{19}{10}\)

\(=-\dfrac{1}{8}-\dfrac{19}{10}=-\dfrac{10}{80}-\dfrac{152}{80}=-\dfrac{162}{80}=-\dfrac{81}{40}\)

\(7,-\dfrac{9}{10}-\left(-\dfrac{7}{18}\right)=-\dfrac{162}{180}-\left(-\dfrac{70}{180}\right)=-\dfrac{92}{180}=-\dfrac{23}{45}\)

\(8,\dfrac{3}{10}-\dfrac{11}{18}=\dfrac{54}{180}-\dfrac{110}{180}=-\dfrac{56}{180}=-\dfrac{14}{45}\)

\(9,\dfrac{3}{5}-\dfrac{5}{6}+\left(-\dfrac{7}{12}\right)=\dfrac{18}{30}-\dfrac{25}{30}+\left(-\dfrac{7}{12}\right)=-\dfrac{7}{30}+\left(-\dfrac{7}{12}\right)\)

\(=-\dfrac{84}{360}+\left(-\dfrac{210}{360}\right)=-\dfrac{294}{360}=-\dfrac{49}{60}\)

\(10,-\dfrac{4}{9}-\dfrac{-5}{6}-\dfrac{3}{8}=-\dfrac{24}{54}-\dfrac{-45}{54}-\dfrac{3}{8}\)

\(=\dfrac{21}{54}-\dfrac{3}{8}=\dfrac{7}{18}-\dfrac{3}{8}=\dfrac{56}{144}-\dfrac{54}{144}=\dfrac{2}{144}=\dfrac{1}{72}\)

`@mt`

Xóa nhanh đi ko CTV xóa giờ

Chia chúng nó ra

Câu dễ vậy bạn cần hỏi

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)

= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)

= \(1-\frac{1}{10}\)

=\(\frac{9}{10}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

=\(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

     \(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

     \(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

      \(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(\frac{10}{11}\)

         A= \(\frac{10}{11}:\frac{2}{3}\)

          A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)

d) giả tương tự câu c kết quả \(\frac{25}{11}\)

tổng đặc biệt đó bạn

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(1-\frac{1}{10}=\frac{9}{10}\)

những câu sau cũng áp dụng như vậy nhé

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\)

\(=1-\frac{1}{11}=\frac{10}{11}\)

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{3}{2}.\left(1-\frac{1}{11}\right)=\frac{3}{2}.\frac{10}{11}=\frac{15}{11}\)

a) \(\frac{7}{11}-\left(\frac{3}{5}+\frac{7}{11}\right)=-\frac{3}{5}\)

b) \(\left(\frac{11}{22}+\frac{5}{11}\right)-\frac{19}{22}=\frac{1}{11}\)

c) \(\frac{2}{9}.\frac{4}{5}+\frac{2}{9}.\frac{14}{5}=\frac{4}{5}\)

d) \(-\frac{3}{2}.\frac{7}{10}-\frac{3}{2}.\frac{1}{10}=-\frac{6}{5}\)

e) \(\left(0,75-1+\frac{1}{4}\right):\left(\frac{1515}{1616}+\frac{1616}{1717}\right)=0\)

\(a,\frac{-1}{9}.\frac{15}{22}.\frac{-9}{25}\)

\(=\frac{-1.15.\left(-9\right)}{9.22.25}\)

\(=\frac{3}{110}\)

\(b,\frac{-2}{7}.\left(\frac{5}{13}-\frac{9}{15}\right)-\frac{2}{7}.\frac{8}{13}\)

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

\(=\frac{-2}{7}.\left(1-\frac{3}{5}\right)\)

\(=\frac{-2}{7}.\frac{2}{5}\)

\(=\frac{-4}{35}\)

\(c,\frac{3}{10}.\left(\frac{-4}{9}+\frac{2}{5}\right)-\frac{3}{10}.\left(\frac{5}{9}-\frac{3}{5}\right)\)

\(=\frac{3}{10}.\left[\left(\frac{-4}{9}+\frac{2}{5}\right)-\left(\frac{5}{9}-\frac{3}{5}\right)\right]\)

\(=\frac{3}{10}.\left(\frac{-4}{9}+\frac{2}{5}-\frac{5}{9}+\frac{3}{5}\right)\)

\(=\frac{3}{10}.\left[\left(\frac{-4}{9}-\frac{5}{9}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\right]\)

\(=\frac{3}{10}.\left(-1+1\right)\)

\(=\frac{3}{10}.0\)

\(=0\)

\(d,\frac{4}{11}-\frac{5}{13}+\frac{7}{11}-\frac{8}{13}\)

\(=\left(\frac{4}{11}+\frac{7}{11}\right)+\left(\frac{-5}{13}-\frac{8}{13}\right)\)

\(=1-1\)

\(=0\)

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