Ghi đầy đủ nha

bn có thể ghi rõ ràng đc ko?

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

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

=\(0+\dfrac{5}{13}\) 

=\(\dfrac{5}{13}\)

a: \(=\left(15-6-\dfrac{13}{18}\right):\dfrac{298}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)

\(=\dfrac{149}{18}\cdot\dfrac{27}{298}-\dfrac{5}{3}=\dfrac{3}{2}-\dfrac{5}{3}=\dfrac{9-10}{6}=\dfrac{-1}{6}\)

b: \(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\dfrac{-22}{15}:\dfrac{11}{2}\)

\(=\dfrac{3}{4}-\dfrac{4}{15}=\dfrac{29}{60}\)

c: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=\dfrac{-7}{9}+\dfrac{7}{9}+5=5\)

d: \(=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot10\cdot\dfrac{1}{5}\cdot\dfrac{3}{4}=1\)

e: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}+\dfrac{-23}{4}=\dfrac{204}{25}\)

Các bài toán này bn bấm máy thì sẽ ra nha.

Bài 7 có quy tắc tính trong ngoặc trước, ngoài ngoặc sau.

7b bn nhóm 2/3 +1/18 lại vào một chỗ và để cho chúng dấu ngoặc

tíc mình nha

a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)

\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)

\(=\dfrac{-1621}{126}\)

b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)

\(=-\dfrac{49}{20}\)

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`