\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)

\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)

-_-

\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)

\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)

\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)

\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)

\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)

\(=7-\dfrac{26}{5}\)

\(=\dfrac{9}{5}\)

\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)

\(=\dfrac{2}{3}+\dfrac{21}{8}\)

\(=\dfrac{79}{24}\)

\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)

\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)

\(=\dfrac{31}{4}:\dfrac{49}{8}\)

\(=\dfrac{62}{49}\)

\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)

Cách 1: Tính giá trị từng biểu thức trong ngoặc

A=

Cách 2: Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp

A =

= (6-5-3) -

= -2 -0 - = - (2 + ) = -2

Lời giải:

Cách 1: Tính giá trị từng biểu thức trong ngoặc

A=

Cách 2: Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp

A =

= (6-5-3) -

= -2 -0 - = - (2 + ) = -2

a) = (\(-\dfrac{141}{20}\)- \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\dfrac{1}{15}\)

    = \(-\dfrac{73}{10}\) : - 5

    = \(\dfrac{73}{50}\)

b) = \(\left(\dfrac{3}{25}-\dfrac{28}{25}\right)\). \(\dfrac{7}{3}\) : \(\left(\dfrac{7}{2}-\dfrac{11}{3}.14\right)\)

    = \(-\dfrac{7}{3}\) . \(-\dfrac{6}{287}\)

    = \(\dfrac{2}{41}\)

\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{10}\)

\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=2\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)

\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)

\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)

\(\Leftrightarrow x=-\dfrac{49}{8}\)

\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)

\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)

\(\Leftrightarrow x=\dfrac{413}{160}\)

a)\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}(21​+1,5)⋅x=51​

2 \cdot x=\dfrac{1}{5}2⋅x=51​

x=\dfrac{1}{5}: 2x=51​:2

 x=\dfrac{1}{10} x=101​
b) \left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}(−153​+x):1312​=261​

-1 \dfrac{3}{5}+x=\dfrac{13}{6} \cdot \dfrac{12}{13}−153​+x=613​⋅1312​
x=2+1 \dfrac{3}{5}x=2+153​

 x=3 \dfrac{3}{5} x=353​
c) $($