a) \(=\left(\left(-\frac{1}{4}-\frac{5}{3}\right)+\frac{7}{33}\right)-\left(-\frac{15}{12}+\frac{6}{11}-\frac{48}{49}\right)\)

\(=\left(-\frac{23}{12}+\frac{7}{33}\right)+\frac{15}{12}-\frac{6}{11}+\frac{48}{49}\)

\(=\left(-\frac{23}{12}+\frac{15}{12}\right)+\left(\frac{9}{33}-\frac{6}{11}\right)+\frac{48}{49}\)

\(=-\frac{2}{3}-\frac{3}{11}+\frac{48}{49}\)

\(=\frac{65}{1617}\)

b) \(=\frac{11}{125}+\left(-\frac{17}{18}+\frac{4}{9}\right)+\left(-\frac{5}{7}+\frac{17}{14}\right)\)

\(=\frac{11}{125}-\frac{1}{2}+\frac{1}{2}\)

\(=\frac{11}{125}\)

tính từng cái là đc

cau a em

A) 11/125+(-17/18+4/9)-(5/7-17/14)

 =11/125+(-17/18+8/18)-(10/14-17/14)

=11/125+(-1/2)-(-1/2)

=11/125

B)(1+2+3+4-3-2-1)-(1/2+1/2+2/3+1/3+3/4+1/4)

 =4-3

=1

à bn lấy trong sách các dạng toán và phương pháp giải hả

a)\(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)

\(=\dfrac{11}{125}-\left(\dfrac{17}{18}-\dfrac{4}{9}\right)-\left(\dfrac{5}{7}-\dfrac{17}{14}\right)\)

\(=\dfrac{11}{125}-\dfrac{5}{18}-\dfrac{-7}{14}=\dfrac{11}{125}-\dfrac{5}{18}+\dfrac{1}{2}\)

\(=\dfrac{11}{125}-\left(\dfrac{5}{18}-\dfrac{1}{2}\right)=\dfrac{11}{125}-\dfrac{-4}{18}=\dfrac{11}{125}+\dfrac{2}{9}\)

a: =35/17-18/17-9/5+4/5

=1-1=0

b: =-7/19(3/17+8/11-1)

=7/19*18/187=126/3553

c: =26/15-11/15-17/3-6/13

=1-6/13-17/3

=7/13-17/3=-200/39

mh biết làm bài này rùi bn có cần mi2h đang cho bn ko?

= 11/125 - [(17/18 - 4/9) + (5/7 - 17/14)]

= 11/125 - (1/2 - 1/2)

= 11/125

\(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}=\dfrac{11}{125}-\left(\dfrac{17}{18}-\dfrac{4}{9}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)=\dfrac{11}{125}-\dfrac{17-4.2}{18}+\dfrac{17-5.2}{14}=\dfrac{11}{125}-\dfrac{9}{18}+\dfrac{7}{14}=\dfrac{11}{125}-\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{11}{125}\)