a: =382-282+531-331

=100+200=300

b: =(7-8)+(9-10)+...+(2009-2010)

=(-1)+(-1)+....+(-1)

=-1002

c: =-(1+2+3+...+2009+2010)

=-2010*2011/2=-2021055

\(a,-\frac{4}{7}+\left(-\frac{2}{3}+\frac{3}{7}\right)=-\frac{4}{7}-\frac{2}{3}+\frac{3}{7}=-\frac{4}{7}+\frac{3}{7}-\frac{2}{3}=-\frac{1}{7}-\frac{2}{3}=-\frac{17}{21}\)

\(b,-\frac{3}{10}\cdot\frac{5}{12}+-\frac{3}{10}\cdot\frac{7}{12}+5\cdot\frac{3}{10}=\frac{3}{10}\cdot-\frac{5}{12}+\frac{3}{10}\cdot-\frac{7}{12}+5\cdot\frac{3}{10}=\frac{3}{10}\cdot\left(-\frac{5}{12}-\frac{7}{12}+5\right)\)

\(=\frac{3}{10}\cdot4=\frac{6}{5}\)

ti ck nha

đúng đó

xin đó

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a, -4/7 + (-2/3 + 3/7 )

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

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

= -1/7 + -14/21

= -3/21 + -14/21

= -17/21

b, -3/10 . 5/12 + -3/10 . 7/12 + 5. - 3/10

-3/10.(5/12 + 7/12 + 60/12)

-3/10.6

-18/10= -9/5

nhìn ko hiểu luôn á rồi sao làm

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

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

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

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

=73 nha bạn

73 cậu nha

Bài 1: 

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

\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)

\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)

Bài 2: 

a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)

\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)

\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)

b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)

Bài 3:

\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)

\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)

Bài 4:

 \(\dfrac{3}{4}-x=1\)

\(\Rightarrow-x=1-\dfrac{3}{4}\)

\(\Rightarrow x=-\dfrac{1}{4}\)

Vậy: \(x=-\dfrac{1}{4}\)

\(x+4=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{5}-4\)

\(\Rightarrow x=-\dfrac{19}{5}\)

Vậy: \(x=-\dfrac{19}{5}\)

\(x-\dfrac{1}{5}=2\)

\(\Rightarrow x=2+\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{11}{5}\)

Vậy: \(x=\dfrac{11}{5}\)

\(x+\dfrac{5}{3}=\dfrac{1}{81}\)

\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)

\(\Rightarrow x=-\dfrac{134}{81}\)

Vậy: \(x=-\dfrac{134}{81}\)

bài1  

a) \(\dfrac{7}{6}-\dfrac{13}{12}+\dfrac{3}{4}\) 

=\(\dfrac{14}{12}-\dfrac{13}{12}+\dfrac{9}{12}\) 

=\(\dfrac{1}{12}+\dfrac{9}{12}\) 

=\(\dfrac{10}{12}=\dfrac{5}{6}\)

bài 1 

b)\(1\dfrac{1}{2}.(\dfrac{-4}{5})\) + \(\dfrac{3}{10}\) 

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

= \(-\dfrac{6}{5}+\dfrac{3}{10}\) 

=\(-\dfrac{12}{10}+\dfrac{3}{10}\) 

=\(-\dfrac{9}{10}\)