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

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

= \(\frac{11}{125}-\left(\frac{17}{18}-\frac{8}{18}\right)-\left(\frac{10}{14}-\frac{17}{14}\right)\)

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

= \(\frac{11}{125}\)

Bài 1:

A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))

A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)

A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)

A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))

A = 1 - 1 - \(\frac{228}{210}\)

A = 0 - \(\frac{128}{210}\)

A = - \(\frac{64}{105}\)

Bài 2:

B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)

B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)

B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))

B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))

B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))

B = 3 - \(\frac{97}{65}\)

B = \(\frac{195}{65}\) - \(\frac{97}{65}\)

B = \(\frac{98}{65}\)

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

\(=\frac{11}{125}-\left(\frac{17}{18}-\frac{8}{18}\right)+\left(\frac{17}{14}-\frac{10}{14}\right)\)

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

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

b) \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)

\(=\left(6-5-3\right)+\left(\frac{7}{3}-\frac{5}{3}-\frac{2}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)

\(=-2+0+\frac{-1}{2}\)

= \(-2-\frac{-1}{2}=-\left(2+\frac{1}{2}\right)=-2\frac{1}{2}\)

ấy chỗ b) thiếu bước đầu mở ngoặc r mới nhóm nhé :))

trong câu hỏi tương tự có một bài giốngđè và được giải rồi, bạn xem thử đi

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

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

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

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

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

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

b)................................đề bài......................................................

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

=       0     +     0     +      0      +         (-1)       +         (-1)      +         (-1)      + 4

=                                            1

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

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

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

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