Question

[•ɦà↭ƙĭềυ↭σαηɦ•]

Tìm x thuộc z ;

a, \(\left(\frac{31}{20}-\frac{26}{45}\right).\frac{-36}{35}< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right).\frac{8}{13}\)

b,\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)

GIÚP MIK VS Ạ, MIK CẦN GẤP LẮM

Answer 1

[nguyen thi hien]

\(a,\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}\)

\(taco:\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)=\frac{35}{36}\cdot\frac{-36}{35}=-1\)

\(\left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}=\frac{13}{8}\cdot\frac{8}{13}=1\)

\(=>x=0\)

\(b,\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}< x< \frac{-1}{2}+2+\frac{5}{2}\)(dau <co dau gach ngang o duoi nha)

\(taco:\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}=\frac{-5}{6}+\frac{8}{3}+\frac{-29}{3}=\frac{-5}{6}+\frac{16}{6}+\frac{-58}{6}=\frac{-47}{6}=-7,8\)

\(\frac{-1}{2}+2+\frac{5}{2}=\frac{3}{2}+\frac{5}{2}=4\)

tu do \(=>x=-7,8;...;0;1;2;3;4\)