rêtrt

A= 5/4

B= 63/52

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

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

-_-

A = 1

a) \(A=\left(\dfrac{1}{3}-\dfrac{8}{15}-\dfrac{1}{7}\right)+\left(\dfrac{2}{3}+\dfrac{-7}{15}+1\dfrac{1}{7}\right)\)

\(A=\dfrac{1}{3}-\dfrac{8}{15}-\dfrac{1}{7}+\dfrac{2}{3}+\dfrac{-7}{15}+1+\dfrac{1}{7}\)

\(A=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{-8}{15}+\dfrac{-7}{15}\right)+\left(\dfrac{-1}{7}+\dfrac{1}{7}\right)+1\)

\(A=1+\left(-1\right)+0+1=\left(1-1\right)+1\)

\(A=1\)

b) \(B=0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)

\(B=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1-\dfrac{1}{4}\)

\(B=\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-1-\dfrac{1}{8}\)

\(B=0+1-1-\dfrac{1}{8}\)

\(B=-\dfrac{1}{8}\)

A=(13+23)−(815+715)+(−17+117)=1−1+1=1A=(13+23)−(815+715)+(−17+117)=1−1+1=1

$=\left(\frac{1}{4}-1-\frac{1}{4}\right)+1-\frac{1}{8}=\frac{-1}{8}$.

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

cách 1:

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

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

= \(\left(\dfrac{18}{3}-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(\dfrac{15}{3}+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(\dfrac{9}{3}-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{16}{3}+\dfrac{1}{2}\right)-\left(\dfrac{20}{3}-\dfrac{3}{2}\right)-\left(\dfrac{2}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{32}{6}+\dfrac{3}{6}\right)-\left(\dfrac{40}{6}-\dfrac{9}{6}\right)-\left(\dfrac{4}{6}+\dfrac{15}{6}\right)\)

= \(\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)

= \(-\dfrac{15}{6}=-\dfrac{5}{2}\)

cách 2:

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

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

= \(\left(-2\right)-0+\dfrac{1}{2}\)

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

1.\(\dfrac{4}{5}\)

2.\(\dfrac{2}{3}\)

3.\(\dfrac{39}{55}\)

a) \(\dfrac{3}{5}\times\dfrac{2}{3}:\dfrac{1}{2}=\dfrac{3}{5}\times\dfrac{2}{3}\times2=\dfrac{6}{15}\times2=\dfrac{12}{15}=\dfrac{4}{5}\)

b) \(\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}\)

c) \(\left(\dfrac{5}{7}+\dfrac{2}{5}\right):\dfrac{11}{7}=\left(\dfrac{25}{35}+\dfrac{14}{35}\right)\cdot\dfrac{7}{11}=\dfrac{39}{35}\cdot\dfrac{7}{11}=\dfrac{273}{385}=\dfrac{39}{55}\)

a) = \(\dfrac{6}{15}\) \(\div\dfrac{1}{2}=\) \(\dfrac{12}{15}\) rút gọn = \(\dfrac{4}{5}\)

b) = \(\dfrac{2}{6}+\dfrac{1}{3}\) = \(\dfrac{6}{18}+\dfrac{6}{18}\) = \(\dfrac{12}{18}\) = \(\dfrac{2}{3}\)

c) = \(\dfrac{5}{7}+\dfrac{2}{5}=\dfrac{25}{35}+\dfrac{14}{35}=\dfrac{39}{35}\) \(\div\dfrac{11}{7}=\dfrac{39}{35}\)

a, \(\begin{array}{l}A = \dfrac{{ - 3}}{{14}} + \dfrac{2}{{13}} + \dfrac{{ - 25}}{{14}} + \dfrac{{ - 15}}{{13}}\\A = \left( {\dfrac{{ - 3}}{{14}} + \dfrac{{ - 25}}{{14}}} \right) + \left( {\dfrac{2}{{13}} + \dfrac{{ - 15}}{{13}}} \right)\\A = \dfrac{{ - 3 + \left( { - 25} \right)}}{{14}} + \dfrac{{2 + \left( { - 15} \right)}}{{13}}\\A = \dfrac{{ - 28}}{{14}} + \dfrac{{ - 13}}{{13}}\\A =  - 2 + (-1)\\A =  - 3\end{array}\)

b, 

Cách 1:

\(\begin{array}{l}B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \left( {\dfrac{5}{3}.\dfrac{7}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}} \right) + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = 0 + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\end{array}\)

Cách 2:

\(B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \dfrac{5}{3}.({\dfrac{7}{{25}} -\dfrac{7}{{25}} + \dfrac{{21}}{{25}}})\\B =  \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\)

Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.

\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)

\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)