b: D=(-3/31-28/31)+(-6/17-11/17)+(1/25-5/25)

=-2-4/25=-54/25

7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)

\(\Rightarrow\dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\)

\(\Rightarrow-2\le x\le2\)

\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)

\(\Rightarrow\dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{6}{12}-\dfrac{2}{12}\right)\)

\(\Rightarrow\dfrac{2}{3}\cdot\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}\cdot\dfrac{4}{12}\)

\(\Rightarrow\dfrac{22}{36}\le\dfrac{x}{18}\le\dfrac{28}{36}\)

\(\Rightarrow\dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\)

\(\Rightarrow x\in\left\{11;12;13;14\right\}\)

8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}.\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}.\dfrac{2}{6}\\ \dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\\ \Rightarrow11\le x\le14\\ \Rightarrow x\in\left\{11;12;13;14\right\}\)

7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\\ \dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\\ \Rightarrow-2\le x\le2\\ \Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

a: \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\)

\(=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{16}\cdot\dfrac{16\cdot17}{2}\)

\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{17}{2}\)

\(=\dfrac{1}{2}\left(2+3+4+...+17\right)\)

\(=\dfrac{1}{2}\cdot152=76\)

b: Sửa đề: \(\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}+\dfrac{11}{3862}\right)\cdot\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\left(\dfrac{2}{193}\cdot\dfrac{193}{17}-\dfrac{3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right):\left[\dfrac{7}{1931}\cdot\dfrac{1931}{25}+\dfrac{11}{3862}\cdot\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\left(\dfrac{2}{17}-\dfrac{3}{34}+\dfrac{33}{34}\right):\left(\dfrac{7}{25}+\dfrac{11}{50}+\dfrac{9}{2}\right)\)

\(=\left(\dfrac{2}{17}+\dfrac{30}{34}\right):\dfrac{14+11+225}{50}\)

\(=1\cdot\dfrac{50}{250}=1\cdot\dfrac{1}{5}=\dfrac{1}{5}\)

c: Sửa đề: \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

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

d: \(\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)

\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)

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

\(=\dfrac{1}{3}+1:\dfrac{3}{2}=1\)

\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)

\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)

\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)

\(=-\dfrac{37}{16}\)

\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)

\(=\dfrac{5}{17}+\dfrac{-3}{17}\)

\(=\dfrac{2}{17}\)

\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)

\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)

\(=2-\dfrac{7}{30}\)

\(=\dfrac{53}{30}\)

\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)

\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)

\(=\dfrac{15}{34}\)

Bài 1:

Tổng của 2 số là

  \(36\times2=72\) 

Số lớn là

  \(72-17=55\) 

Bài 2:

a) \(4567+y\div34=10987\) 

                 \(y\div34=10987-4567\) 

                 \(y\div34=6420\) 

                         \(y=6420\times34\) 

                         \(y=218280\) 

b) \(\dfrac{4}{3}+\dfrac{1}{2}\div y=2\) 

             \(\dfrac{1}{2}\div y=2-\dfrac{4}{3}\) 

              \(\dfrac{1}{2}\div y=\dfrac{2}{3}\) 

                      \(y=\dfrac{1}{2}\div\dfrac{2}{3}\) 

                      \(y=\dfrac{3}{4}\) 

Bài 3:

a) \(\dfrac{2}{5}\times\dfrac{2}{5}+\dfrac{9}{8}\div3=\dfrac{4}{25}+\dfrac{9}{8}\times\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}=\dfrac{107}{200}\) 

b) \(2-\left(\dfrac{1}{7}\times4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{25}{21}\)

Bài 1 : Gọi a là số lớn, b là số bé, theo đề bài ta có :

(a+b):2=36⇒a+b=72

mà b=17

Nên a=72-17=55

Bài 2 :

a) 4567+y:34=10987

⇒ y:34=10987-4567

⇒ y:34=6420

⇒ y=6420x34

⇒ y=218280

b) \(\dfrac{4}{3}+\dfrac{1}{2}:y=2\)

\(\Rightarrow\dfrac{1}{2}:y=2-\dfrac{4}{3}\)

\(\Rightarrow\dfrac{1}{2}:y=\dfrac{2}{3}\)

\(\Rightarrow y=\dfrac{1}{2}:\dfrac{2}{3}\)

\(\Rightarrow y=\dfrac{1}{2}x\dfrac{3}{2}\)

\(\Rightarrow y=\dfrac{3}{4}\)

Bài 3 :

\(\dfrac{2}{5}x\dfrac{2}{5}+\dfrac{9}{8}:3=\dfrac{4}{25}+\dfrac{9}{8}x\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}\)

= \(\dfrac{4x8}{25x8}+\dfrac{25x3}{25x8}=\dfrac{32}{200}+\dfrac{75}{200}=\dfrac{107}{200}\)

\(2-\left(\dfrac{1}{7}x4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\left(\dfrac{12}{21}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{42}{21}-\dfrac{17}{21}=\dfrac{25}{21}\)

o: \(\dfrac{\left(-1\right)^6\cdot3^5\cdot4^3}{9^2\cdot2^5}=\dfrac{3^5\cdot2^6}{2^5\cdot3^4}=\dfrac{3^5}{3^4}\cdot\dfrac{2^6}{2^5}=3\cdot2=6\)

s: \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)

\(=\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}{\dfrac{3}{2}\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}\)

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

t: \(\sqrt{\dfrac{4}{9}}-\dfrac{1}{2}:\left|-\dfrac{2}{3}\right|\)

\(=\dfrac{2}{3}-\dfrac{1}{2}:\dfrac{2}{3}\)

\(=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8-9}{12}=-\dfrac{1}{12}\)

a,=\(\dfrac{\left(2-\dfrac{1}{3}+\dfrac{1}{4}\right).12}{\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right).12}\)+\(\dfrac{\left(\dfrac{3}{5}-\dfrac{1}{4}+\dfrac{1}{2}\right).20}{\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{2}{5}\right).20}\)

=\(\dfrac{24-4+3}{24+2-3}\) +\(\dfrac{12-5+10}{10+15-8}\)(nhân từng số hạng với 12;20)

=\(\dfrac{23}{23}\)+\(\dfrac{17}{17}\) =1+1=2

b,=\(\dfrac{5.\left(\dfrac{1}{79}\right)+5.\left(\dfrac{1}{83}\right)+\dfrac{1}{17}}{17.\left(\dfrac{1}{79}\right)+17.\left(\dfrac{1}{83}\right)+\dfrac{1}{5}}\)=\(\dfrac{5.\left(\dfrac{1}{79}+\dfrac{1}{83}\right)+\dfrac{1}{17}}{17.\left(\dfrac{1}{79}+\dfrac{1}{83}\right)+\dfrac{1}{5}}\)

\(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\cdot\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{1-10}{6}-\dfrac{20}{3}\cdot\dfrac{8-5}{20}+\dfrac{2}{3}\cdot\dfrac{12-45}{10}\)

\(=\dfrac{-1}{2}\cdot\dfrac{-9}{6}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{66}{30}=\dfrac{-1}{4}-\dfrac{11}{5}=\dfrac{-5-44}{20}=-\dfrac{49}{20}\)