a: \(=\dfrac{7+12-6}{13}=1\)

b: \(=\dfrac{13}{10}\cdot\dfrac{6-26}{13}=\dfrac{-20}{10}=-2\)

c: \(=\dfrac{3}{4}\cdot2-\dfrac{5}{2}\cdot\dfrac{-4}{3}=\dfrac{3}{2}+\dfrac{20}{6}=\dfrac{3}{2}+\dfrac{10}{3}=\dfrac{29}{6}\)

d: \(=\dfrac{3}{8}\cdot\dfrac{8}{5}+\dfrac{3}{5}\cdot\dfrac{2}{7}+\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{5}+\dfrac{3}{5}=\dfrac{6}{5}\)

Bài 4. Trên tia Ox lấy điểm A và B sao cho OA = 8cm, OB = 12cm.

a, Điểm A có nằm giữa hai điểm O và B không? Vì sao? Tính AB.

b, Gọi M, N lần lượt là trung điểm của OA, OB. Điểm M có nằm giữa hai điểm O và N không? Vì sao? Tính MN.

c, Điểm N có là trung điểm của đoạn thẳng AM không? Vì sao? 
cho mìn hỏi câu b nhoa

c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)

\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)

\(=1-2=-1\)

Giải:

a)-1/12+4/3=-1/12+16/12=15/12=5/4

b)(-4/14-3/15)-(1/5-20/35-(-1)).7

=-17/35-22/35.7

=-17/35-22/5

=-171/35

c)3/5+-5/20+30/75+-7/4

=3/5+-1/4+2/5+-7/4

=(3/5+2/5)+(-1/4+-7/4)

=1+-2

=-1

d)5/6.-12/14+7/13

=-5/7+7/13

=-16/91

e)2/-9-5/-36-1/4

=-1/12-1/4

=-1/3

f)2/23+-5/12+7/18+21/23+-7/12

=(2/23+21/23)+(-5/12+-7/12)+7/18

=1+-1+7/18

=7/18

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ai lm tr mình cho tích

a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)

\(\begin{array}{l} = \frac{2}{5}.\left( {\frac{{ - 3}}{7} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\left( {\frac{{ - 6}}{{14}} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\frac{{ - 11}}{{14}} - \frac{{18}}{{35}} = \frac{{ - 11}}{{35}} - \frac{{18}}{{35}} =  \frac{{ -29}}{{35}}\end{array}\)

b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\)

\(\begin{array}{l} = \left( {\frac{{2.11.4}}{{3.11.4}} - \frac{{5.3.4}}{{11.3.4}} + \frac{{1.3.11}}{{4.3.11}}} \right):\left( {\frac{11.12}{11.12} + \frac{{5.11}}{{12.11}} - \frac{{7.12}}{{11.12}}} \right)\\ = \left( {\frac{{88 - 60 + 33}}{{121}}} \right):\left( { \frac{{121+55 - 84}}{{121}}} \right)\\ = \frac{{61}}{{121}}:\frac{{92}}{{121}} = \frac{{61}}{{121}}.\frac{{121}}{{92}}= \frac{{61}}{{92}}\end{array}\)

c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)

\( = \left( { - 24,2} \right).\left( { - 3,2} \right) = 77,44\)

d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)

\(\begin{array}{l} = \left( { - 25,4} \right).\left( {62,1 - 16,8} \right):12,7\\ = \left( { - 25,4} \right).45,3:12,7\\ = \left( { - 25,4} \right):12,7.45,3\\ =  (- 2).45,3 =  - 90,6\end{array}\)

a: \(=\dfrac{2}{5}\cdot\left(-\dfrac{3}{7}-\dfrac{5}{14}\right)-\dfrac{18}{35}\)

\(=\dfrac{2}{5}\cdot\dfrac{-6-5}{14}-\dfrac{18}{35}\)

\(=\dfrac{2}{5}\cdot\dfrac{-11}{14}-\dfrac{18}{35}=-\dfrac{22}{70}-\dfrac{18}{35}=\dfrac{-58}{70}=-\dfrac{29}{35}\)

b: \(=\dfrac{88-60+33}{132}:\dfrac{132+55-84}{132}\)

\(=\dfrac{61}{132}\cdot\dfrac{132}{103}=\dfrac{61}{103}\)

c: \(=-24.2\cdot\left(-3.2\right)=24.2\cdot3.2=77.44\)

d: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)

b: \(B=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{3}{8}+\dfrac{6}{8}+\dfrac{-6}{11}-\dfrac{5}{11}=-2-1+\dfrac{9}{8}=\dfrac{9}{8}-3=-\dfrac{15}{8}\)

c: \(C=\left(\dfrac{4}{3}+\dfrac{7}{3}+\dfrac{1}{3}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)=4+1=5\)

d: \(D=\dfrac{4}{19}\left(\dfrac{-5}{6}-\dfrac{7}{12}\right)-\dfrac{40}{57}\)

\(=\dfrac{4}{19}\cdot\dfrac{-17}{12}-\dfrac{40}{57}=-1\)

e: \(E=\dfrac{1}{3}\left(\dfrac{4}{5}-\dfrac{9}{5}\right)+\dfrac{2}{3}=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)