Bài 6:
Để x là số hữu tỉ thì: \(a\in Z\)
a) Để x là số dương thì:
\(x=\dfrac{a-5}{2}>0=>a-5>0\\ =>a>5\)
b) Để x là số âm thì:
\(x=\dfrac{a-5}{2}< 0=>a-5< 0\\ =>a< 5\)
c) Để x không phải là số âm không phải là số dơng thì:
\(x=\dfrac{a-5}{2}=0=>a-5=0\\ =>a=5\)
Bài 7:
Ta có: \(x=\dfrac{a-5}{a}=\dfrac{a}{a}-\dfrac{5}{a}=1-\dfrac{5}{a}\)
Để x là số nguyên thì: 5 ⋮ a
=> a ∈ Ư(5) = {1; -1; 5; -5}
Bài 4:
a: Vì -16<-14<-12<-11<-9<-3<-1
nên \(-\dfrac{16}{17}< -\dfrac{14}{17}< -\dfrac{12}{17}< -\dfrac{11}{17}< -\dfrac{9}{17}< -\dfrac{3}{17}< -\dfrac{1}{17}\)
b: 2<3<4<7<8<9<11
=>\(\dfrac{5}{2}>\dfrac{5}{3}>\dfrac{5}{4}>\dfrac{5}{7}>\dfrac{5}{8}>\dfrac{5}{9}>\dfrac{5}{11}\)
=>\(-\dfrac{5}{2}< -\dfrac{5}{3}< -\dfrac{5}{4}< -\dfrac{5}{7}< -\dfrac{5}{8}< -\dfrac{5}{9}< -\dfrac{5}{11}\)
c: 37>33
=>\(\dfrac{14}{37}< \dfrac{14}{33}\)
=>\(-\dfrac{14}{33}< -\dfrac{14}{37}< 0\left(1\right)\)
\(\dfrac{17}{20}=\dfrac{17\cdot19}{20\cdot19}=\dfrac{323}{380};\dfrac{18}{19}=\dfrac{18\cdot20}{19\cdot20}=\dfrac{360}{380}\)
mà 323<360
nên \(0< \dfrac{17}{20}< \dfrac{18}{19}< 1\)
mà \(1=\dfrac{3}{3}< \dfrac{4}{3}\)
nên \(0< \dfrac{17}{20}< \dfrac{18}{19}< \dfrac{4}{3}\left(2\right)\)
Từ (1),(2) suy ra \(-\dfrac{14}{33}< -\dfrac{14}{37}< 0< \dfrac{17}{20}< \dfrac{18}{19}< \dfrac{4}{3}\)
Bài 5:
\(-\dfrac{1}{3}=\dfrac{-150}{450};\dfrac{4}{5}=\dfrac{4\cdot90}{5\cdot90}=\dfrac{360}{450}\)
=>Ba số hữu tỉ khác nhau lớn hơn -1/3 và nhỏ hơn 4/5 là:
\(-\dfrac{100}{450}=-\dfrac{2}{9};\dfrac{150}{450}=\dfrac{1}{3};\dfrac{270}{450}=\dfrac{3}{5}\)
Bài 8:
a: \(\dfrac{1234}{1235}=1-\dfrac{1}{1235};\dfrac{4319}{4320}=1-\dfrac{1}{4320}\)
1235<4320
=>\(\dfrac{1}{1235}>\dfrac{1}{4320}\)
=>\(-\dfrac{1}{1235}< -\dfrac{1}{4320}\)
=>\(-\dfrac{1}{1235}+1< -\dfrac{1}{4320}+1\)
=>\(\dfrac{1234}{1235}< \dfrac{4319}{4320}\)
b: \(\dfrac{-1234}{1244}=-1+\dfrac{10}{1244};\dfrac{-4321}{4331}=-1+\dfrac{10}{4331}\)
mà \(\dfrac{10}{1244}>\dfrac{10}{4331}\)
nên \(-\dfrac{1234}{1244}>\dfrac{-4321}{4331}\)
c: \(\dfrac{-31}{-32}=\dfrac{31}{32}=\dfrac{31\cdot32327}{32\cdot32327}=\dfrac{1002137}{1034464};\dfrac{31317}{32327}=\dfrac{31317\cdot32}{32327\cdot32}=\dfrac{1002144}{1034464}\)
mà 1002137<1002144
nên \(\dfrac{-31}{-32}< \dfrac{31317}{32327}\)
d: \(\dfrac{3246}{-3247}=\dfrac{-3246}{3247}>-1;-1=\dfrac{-45983}{45983}>\dfrac{-45984}{45983}\)
Do đó: \(\dfrac{3246}{-3247}>\dfrac{-45984}{45983}\)
e: \(\dfrac{22}{67}< \dfrac{22}{66}=\dfrac{1}{3};\dfrac{51}{152}>\dfrac{51}{153}=\dfrac{1}{3}\)
Do đó: \(\dfrac{22}{67}< \dfrac{51}{152}\)
=>\(\dfrac{22}{-67}>\dfrac{51}{-152}\)
f: \(\dfrac{-18}{91}< -\dfrac{18}{90}=-\dfrac{1}{5}\)
\(\dfrac{-1}{5}=\dfrac{-23}{115}< -\dfrac{23}{114}\)
Do đó: \(-\dfrac{18}{91}< -\dfrac{23}{114}\)
a: \(\dfrac{1234}{1235}=1-\dfrac{1}{1235};\dfrac{4319}{4320}=1-\dfrac{1}{4320}\)
1235<4320
=>\(\dfrac{1}{1235}>\dfrac{1}{4320}\)
=>\(-\dfrac{1}{1235}< -\dfrac{1}{4320}\)
=>\(-\dfrac{1}{1235}+1< -\dfrac{1}{4320}+1\)
=>\(\dfrac{1234}{1235}< \dfrac{4319}{4320}\)
b: \(\dfrac{-1234}{1244}=-1+\dfrac{10}{1244};\dfrac{-4321}{4331}=-1+\dfrac{10}{4331}\)
mà \(\dfrac{10}{1244}>\dfrac{10}{4331}\)
nên \(-\dfrac{1234}{1244}>\dfrac{-4321}{4331}\)
c: \(\dfrac{-31}{-32}=\dfrac{31}{32}=\dfrac{31\cdot32327}{32\cdot32327}=\dfrac{1002137}{1034464};\dfrac{31317}{32327}=\dfrac{31317\cdot32}{32327\cdot32}=\dfrac{1002144}{1034464}\)
mà 1002137<1002144
nên \(\dfrac{-31}{-32}< \dfrac{31317}{32327}\)
d: \(\dfrac{3246}{-3247}=\dfrac{-3246}{3247}>-1;-1=\dfrac{-45983}{45983}>\dfrac{-45984}{45983}\)
Do đó: \(\dfrac{3246}{-3247}>\dfrac{-45984}{45983}\)
e: \(\dfrac{22}{67}< \dfrac{22}{66}=\dfrac{1}{3};\dfrac{51}{152}>\dfrac{51}{153}=\dfrac{1}{3}\)
Do đó: \(\dfrac{22}{67}< \dfrac{51}{152}\)
=>\(\dfrac{22}{-67}>\dfrac{51}{-152}\)
f: \(\dfrac{-18}{91}< -\dfrac{18}{90}=-\dfrac{1}{5}\)
\(\dfrac{-1}{5}=\dfrac{-23}{115}< -\dfrac{23}{114}\)
Do đó: \(-\dfrac{18}{91}< -\dfrac{23}{114}\)
