\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)

\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)

\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)

\(\Rightarrow-1\le x< 6\)

\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)

Bài b tương tự

bạn ơi bạn giải câu b được ko. mk ko biết làm câu b

\(A=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(-\frac{20}{31}-\frac{11}{31}\right)+-\frac{4}{9}\)

\(A=1+\left(-1\right)+\frac{-4}{9}\)

\(A=0+\frac{-4}{9}=\frac{-4}{9}\)

\(B=\left(\frac{-3}{7}+\frac{-4}{7}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{-2}{3}\)

\(B=-1+1+\frac{-2}{3}\)

\(B=\frac{-2}{3}\)

A=\(\frac{-4}{9}\)

B=\(\frac{-2}{3}\)

\(A=\frac{5}{17}+-\frac{4}{9}-\frac{20}{31}+\frac{12}{17}-\frac{11}{31}\)

  \(=\left(\frac{5}{17}+\frac{12}{17}\right)+\frac{-4}{9}-\left(\frac{20}{31}+\frac{11}{31}\right)\)

   \(=\frac{17}{17}+\frac{-4}{9}-\frac{31}{31}\)  

  \(=1+\frac{-4}{9}-1=-\frac{4}{9}\)

Ps : Câu tiếp bạn lm giống vậy. Thấy cùng mẫu thì cko nó vào riêng dấu ngoặc nhé! 

a) \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)< \(x\) < \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)

Ta có: \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)

=\(\frac{-40}{15}+\frac{21}{15}+\frac{-71}{15}\)

=\(\frac{-90}{15}\)

=\(-6\)

Ta có: \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)

=\(\frac{-26}{14}+\frac{19}{14}+\frac{-49}{14}\)

=\(\frac{-56}{14}\)

=\(-4\)

=> \(-6\)< \(x\)<\(-4\)

=> \(x=-5\)

b)\(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)< \(\frac{x}{9}\)<\(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)

Ta có: \(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)

=\(\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}+\frac{-11}{31}\right)+\frac{-4}{9}\)

=\(1+\left(-1\right)+\frac{-4}{9}\)

=\(0+\frac{-4}{9}\)

=\(\frac{-4}{9}\)

Ta có: \(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)

=\(\frac{-3}{7}+\frac{7}{15}+\frac{-4}{7}+\frac{8}{15}+\frac{2}{3}\)

=\(\left(\frac{-3}{7}+\frac{-4}{7}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{2}{3}\)

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

=\(0+\frac{2}{3}\)

=\(\frac{2}{3}\)

=> \(\frac{-4}{9}\)< \(\frac{x}{9}\)<\(\frac{2}{3}\)

=

=> \(\frac{-4}{9}\)<\(\frac{x}{9}\)<\(\frac{6}{9}\)

=> \(-4\)< \(x\)<\(6\)

=>\(x\in\left\{-3;-2;-1;0;1;2;3;4;5\right\}\)

=>(5/17+12/17)+(-20/31-11/31)-4/9<=x/9<=(-3/7-4/7)+(7/15+8/15)+2/3

=>-4/9<=x/9<=6/9

=>-4<=x<=6

hay \(x\in\left\{-4;-3;-2;-1;0;...;6\right\}\)

\(A=\frac{3}{2}\times\left(\frac{1}{13\times11}+\frac{1}{13\times15}+\frac{1}{15\times17}+.....+\frac{1}{97\times99}\right)\)

\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+......+\frac{1}{97}-\frac{1}{99}\right)\)

\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{99}\right)\)

\(A=\frac{3}{2}\times\frac{8}{99}\)

\(A=\frac{4}{33}\)

b] \(\frac{A}{5}=\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\)

\(\frac{A}{5}=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\)

\(\frac{A}{5}=\frac{1}{31}-\frac{1}{57}\)

\(\Rightarrow A=5\left(\frac{1}{31}-\frac{1}{57}\right)=\frac{130}{1767}\)

c] Ta đặt \(\left(8n+5,6n+4\right)=d\)

\(\Rightarrow\frac{8n+5\div d}{6n+4\div d}\Rightarrow4\times\left(6n+4\right)-3\times\left(8n+5\right)=\left(24n+16\right)-\left(24n+15\right):d\)\(\Rightarrow d=1\)

Vậy \(\frac{8n+5}{6n+4}\)là phân số tối giản

có rất nhiều câu dễ ở trong đề sao bạn Ko thử làm đi rồi câu nào khó lại hỏi

\(A=17\frac{2}{31}-\left(\frac{15}{17}+6\frac{2}{31}\right)=\left(17\frac{2}{31}-6\frac{2}{31}\right)-\frac{15}{17}=11-\frac{15}{17}=10+\left(1-\frac{15}{17}\right)=10\frac{2}{17}\)

\(B=\left(31\frac{6}{13}-36\frac{6}{13}\right)+5\frac{9}{41}=-5+5\frac{9}{41}=\frac{9}{41}\)

C=\(\left(27\frac{51}{59}-7\frac{51}{59}\right)+\frac{1}{3}=20+\frac{1}{3}=20\frac{1}{3}\)

\(D=\left(13\frac{29}{31}-2\frac{28}{31}\right)+\left(4-3\frac{7}{8}\right)=11\frac{1}{31}+\frac{1}{8}=11\frac{8+31}{31.8}=11\frac{39}{248}\)