a: \(B=\frac12-\left\lbrack\frac38+\left(-\frac74\right)\right\rbrack\)

\(=\frac12-\frac38+\frac74\)

\(=\frac48-\frac38+\frac{14}{8}=\frac{15}{8}\)

b: \(-\frac{4}{12}-\left(-\frac{13}{39}-0,25\right)+0,75\)

\(=-\frac13+\frac13+0,25+0,75\)

=0,25+0,75

=1

c: \(\frac12-\frac13+\frac{1}{23}+\frac16\)

\(=\left(\frac12-\frac13+\frac16\right)+\frac{1}{23}\)

\(=\left(\frac16+\frac16\right)+\frac{1}{23}=\frac13+\frac{1}{23}=\frac{26}{69}\)

d: \(\left(-\frac{13}{7}-\frac49\right)-\left(-\frac{10}{7}-\frac49\right)\)

\(=-\frac{13}{7}-\frac49+\frac{10}{7}+\frac49\)

\(=-\frac{13}{7}+\frac{10}{7}=-\frac37\)

e: \(\left(\frac78-\frac52+\frac47\right)-\left(-\frac37+1-\frac{13}{8}\right)\)

\(=\frac78-\frac52+\frac47+\frac37-1+\frac{13}{8}\)

\(=\frac{20}{8}-\frac52=0\)

f: \(-\frac37+\left(3-\frac34\right)-\left(2,25-\frac{10}{7}\right)\)

\(=-\frac37+3-\frac34-2,25+\frac{10}{7}\)

\(=\frac77+0,75-\frac34\)

=1
g: \(\left(\frac53-\frac37+9\right)-\left(2+\frac57-\frac23\right)+\left(\frac87-\frac43-10\right)\)

\(=\frac53-\frac37+9-2-\frac57+\frac23+\frac87-\frac43-10\)

\(=\left(\frac53+\frac23-\frac43\right)+\left(-\frac37-\frac57+\frac87\right)+\left(9-2-10\right)\)

\(=\frac33+7-10=1-3=-2\)

a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)

= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)

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

= (-1) + \(\dfrac{3}{4}\)

= \(\dfrac{-4}{4}+\dfrac{3}{4}\)

= \(\dfrac{-1}{4}\)

b; 0,5  + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)

= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))

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

= 1 + 1 

= 2

a) A = \(9\frac{3}{8}-\left(2\frac{3}{5}+2\frac{3}{8}\right)=9\frac{3}{8}-2\frac{3}{5}-2\frac{3}{8}=\left(9\frac{3}{8}-2\frac{3}{8}\right)-2\frac{3}{5}=7-\frac{13}{5}=\frac{22}{5}\)

b) B = \(\left(15\frac{3}{5}+5\frac{3}{4}\right)-8\frac{3}{5}=15\frac{3}{5}+5\frac{3}{4}-8\frac{3}{5}=\left(15\frac{3}{5}-8\frac{3}{5}\right)+5\frac{3}{4}=7+\frac{23}{4}=\frac{51}{4}\)

c) C = \(17\frac{1}{4}-\left(2\frac{3}{7}+7\frac{1}{4}\right)=17\frac{1}{4}-2\frac{3}{7}-7\frac{1}{4}=\left(17\frac{1}{4}-7\frac{1}{4}\right)-2\frac{3}{7}=10-\frac{17}{7}=\frac{53}{7}\)

d) D = \(\left(11\frac{5}{17}+3\frac{5}{7}\right)-4\frac{5}{17}=11\frac{5}{17}+3\frac{5}{7}-4\frac{5}{17}=\left(11\frac{5}{17}-4\frac{5}{17}\right)+3\frac{5}{7}=7+\frac{26}{7}=\frac{75}{7}\)

Là sao!!!!!???????!!!!!???

nhìn ko hiểu luôn á rồi sao làm