\(\dfrac{2}{3}+\dfrac{5}{6}+\dfrac{9}{10}+\dfrac{14}{15}+\dfrac{20}{21}+\dfrac{27}{28}+\dfrac{35}{36}+\dfrac{44}{45}\\ =\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{10}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{21}\right)+\left(1-\dfrac{1}{28}\right)+\left(1-\dfrac{1}{36}\right)+\left(1-\dfrac{1}{45}\right)\\ =8-\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\right)\\ =8-\left(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\\ =8-\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+\dfrac{2}{9.10}\right)\\ =8-2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\\ =8-2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =8-2\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=8-2.\dfrac{2}{5}=8-\dfrac{4}{5}=\dfrac{36}{5}\)

he nho

\(e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)

\(=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}\)

\(=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

\(f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)

\(=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}\)

\(=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}\)

\(g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)

\(=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)\)

\(=\dfrac{5}{9}\cdot1=\dfrac{5}{9}\)

a: =-10/21

b: =-5/27

c: =-20/20=-1

d: =-15/105=-1/7

e: =-63/63=-1

\(\dfrac{10}{21}\)

\(\dfrac{5}{27}\)

\(1\)

\(\dfrac{1}{7}\)

\(1\)

\(=\dfrac{85}{18}:\dfrac{85}{9}-\dfrac{136}{45}:\dfrac{136}{15}=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\)

Để tính tổng của biểu thức này, chúng ta cần thực hiện các phép cộng và trừ theo thứ tự từ trái sang phải.

\[4 + \frac{5}{6} - \frac{1}{9} \times \frac{1}{10} - \frac{7}{12} + \frac{1}{36} - 3 - \frac{1}{5} + \frac{1}{3} - \frac{1}{9} \times \frac{9}{5} + 1 - \frac{1}{3}\]

Đầu tiên, chúng ta sẽ làm các phép tính liên quan đến phân số:

\[= 4 + \frac{5}{6} - \frac{1}{90} - \frac{7}{12} + \frac{1}{36} - 3 - \frac{1}{5} + \frac{1}{3} - \frac{1}{5} + 1 - \frac{1}{3}\]

Tiếp theo, chúng ta sẽ tổng hợp các phân số:

\[= 4 + \frac{5}{6} - \frac{1}{90} - \frac{35}{90} + \frac{5}{180} - 3 - \frac{18}{90} + \frac{60}{180} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{1}{90} - \frac{35}{90} + \frac{5}{180} - 3 - \frac{2}{10} + \frac{10}{30} - \frac{2}{10} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{1}{90} - \frac{35}{90} + \frac{5}{180} - 3 - \frac{1}{5} + \frac{1}{3} - \frac{1}{5} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{36 + 35}{90} + \frac{5}{180} - 3 - \frac{1}{5} + \frac{2}{6} - \frac{1}{5} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - 3 - \frac{1}{5} + \frac{1}{3} - \frac{1}{5} + 1 - \frac{1}{3}\]

Tiếp theo, chúng ta sẽ tính tổng các số nguyên:

\[= 4 - 3 + 1\]

Cuối cùng, chúng ta sẽ tổng hợp các phân số:

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{1}{5} + \frac{1}{3} - \frac{1}{5} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{30}{90}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6}

2:

a: 2/9-x=-5/9

=>x=2/9+5/9=7/9

b: x-7/13=1/2

=>x=1/2+7/13=27/26

câu a

\(\dfrac{7}{4}+\dfrac{3}{2}+\dfrac{-9}{16}\\ =\dfrac{28}{16}+\dfrac{24}{16}-\dfrac{9}{16}=\dfrac{43}{16}\)

câu b

\(-\dfrac{2}{7}+\dfrac{3}{5}+\dfrac{9}{7}+\dfrac{-18}{5}\\ =-\dfrac{10}{35}+\dfrac{21}{35}+\dfrac{45}{35}-\dfrac{126}{35}\\ =-\dfrac{70}{35}=-2\)

câu c

\(-\dfrac{5}{13}+\dfrac{11}{10}-\dfrac{-9}{10}+\dfrac{-8}{13}\\ =-\dfrac{5}{13}+\dfrac{11}{10}+\dfrac{9}{10}-\dfrac{8}{13}\\ =-\dfrac{50}{130}+\dfrac{143}{130}+\dfrac{117}{130}-\dfrac{80}{130}\\ =\dfrac{130}{130}=1\)

bài 2

câu a

\(\dfrac{2}{9}-x=-\dfrac{5}{9}\\ x=\dfrac{2}{9}-\dfrac{-5}{9}\\ x=\dfrac{7}{9}\)

câu b

\(x+\dfrac{-7}{13}=\dfrac{1}{2}\\ x=\dfrac{1}{2}-\dfrac{-7}{13}\\ x=\dfrac{13}{26}+\dfrac{14}{26}\\ x=\dfrac{17}{26}\)

a) Ta có: \(\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-1}{5}\)

\(\Leftrightarrow\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-5}{25}=\dfrac{2}{25}\)

hay \(x=\dfrac{6}{25}\)

Vậy: \(x=\dfrac{6}{25}\)

b) Ta có: \(\dfrac{4}{9}+\dfrac{x}{5}=\dfrac{5}{11}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{45}{99}-\dfrac{44}{99}=\dfrac{1}{99}\)

hay \(x=\dfrac{5}{99}\)

Vậy: \(x=\dfrac{5}{99}\)

\(=\dfrac{251}{18}\)

251/18

Chọn D

\(a,x=\dfrac{1}{5}+\dfrac{-3}{7}\)

   \(x=\dfrac{7}{35}+\dfrac{-15}{35}\)

   \(x=-\dfrac{8}{35}\)

\(b,\dfrac{3}{5}-\dfrac{4}{7}:x=\dfrac{-9}{10}\)

           \(\dfrac{4}{7}:x=\dfrac{3}{5}-\dfrac{-9}{10}\)

           \(\dfrac{4}{7}:x=\dfrac{3}{2}\)

                 \(x=\dfrac{4}{7}:\dfrac{3}{2}\)

                 \(x=\dfrac{4}{7}\times\dfrac{2}{3}\)

                 \(x=\dfrac{8}{21}\)

\(c,x-\left(\dfrac{-3}{4}\right)=\dfrac{-2}{3}-\dfrac{1}{2}\)

   \(x+\dfrac{3}{4}=\dfrac{-4}{6}-\dfrac{3}{6}\)

   \(x+\dfrac{3}{4}=-\dfrac{7}{6}\)

           \(x=-\dfrac{7}{6}-\dfrac{3}{4}\)

           \(x=-\dfrac{23}{12}\)

\(d,\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)

    \(\dfrac{-5}{9}-x=\dfrac{6}{18}+\dfrac{7}{18}\)

     \(\dfrac{-5}{9}-x=\dfrac{13}{18}\)

                \(x=\dfrac{-5}{9}-\dfrac{13}{18}\)

                \(x=\dfrac{-10}{18}-\dfrac{13}{18}\)

                \(x=-\dfrac{23}{18}\)

\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)

a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)

\(=\dfrac{2}{3}+\dfrac{2}{7}\)

\(=\dfrac{14}{21}+\dfrac{6}{21}\)

\(=\dfrac{20}{21}\)