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Bài 7:
7.1: I là trung điểm của AB
=>\(AB=2\cdot IA=4\left(cm\right)\)
7.2:
C nằm giữa A và B
=>AC+CB=AB
=>CB=10-8=2(cm)
C là trung điểm của NB
=>NC=CB=2cm
C là trung điểm của NB
=>\(NB=2\cdot NC=2\cdot2=4\left(cm\right)\)
Bài 6:
a: \(\dfrac{4}{5}=\dfrac{4\cdot6}{5\cdot6}=\dfrac{24}{30}\)
\(\dfrac{8}{15}=\dfrac{8\cdot2}{15\cdot2}=\dfrac{16}{30}\)
\(-\dfrac{3}{2}=\dfrac{-3\cdot15}{2\cdot15}=-\dfrac{45}{30}\)
b: \(2=\dfrac{2\cdot45}{45}=\dfrac{90}{45}\)
\(\dfrac{-10}{5}=\dfrac{-10\cdot9}{5\cdot9}=\dfrac{-90}{45}\)
\(\dfrac{7}{-9}=\dfrac{-7}{9}=\dfrac{-7\cdot5}{9\cdot5}=\dfrac{-35}{45}\)
c: \(\dfrac{3}{-2}=\dfrac{-3}{2}=\dfrac{-3\cdot6}{2\cdot6}=\dfrac{-18}{12}\)
\(\dfrac{5}{-6}=\dfrac{-5}{6}=\dfrac{-5\cdot2}{6\cdot2}=\dfrac{-10}{12}\)
\(\dfrac{-6}{4}=\dfrac{-6\cdot3}{4\cdot3}=\dfrac{-18}{12}\)
d: \(-\dfrac{1}{2}=\dfrac{-1\cdot15}{2\cdot15}=\dfrac{-15}{30}\)
\(\dfrac{4}{3}=\dfrac{4\cdot10}{3\cdot10}=\dfrac{40}{30}\)
\(\dfrac{6}{-5}=\dfrac{-6}{5}=\dfrac{-6\cdot6}{5\cdot6}=\dfrac{-36}{30}\)
bài 5:
a: \(\dfrac{3}{4}=\dfrac{9}{12};\dfrac{-3}{12}=\dfrac{-3}{12};\dfrac{-2}{3}=-\dfrac{8}{12};\dfrac{-1}{-6}=\dfrac{1}{6}=\dfrac{2}{12}\)
mà -8<-3<2<9
nên \(-\dfrac{8}{12}< -\dfrac{3}{12}< \dfrac{2}{12}< \dfrac{9}{12}\)
=>\(\dfrac{-2}{3}< \dfrac{-3}{12}< \dfrac{-1}{-6}< \dfrac{3}{4}\)
b: Ta có: \(\dfrac{-7}{9}=\dfrac{-28}{36};\dfrac{-1}{3}=\dfrac{-12}{36};-1=-\dfrac{36}{36}\)
mà -36<-28<-12
nên \(-1< -\dfrac{28}{36}< -\dfrac{12}{36}\)
=>\(-1< \dfrac{-7}{9}< -\dfrac{1}{3}< 0\)
\(\dfrac{5}{12}=\dfrac{15}{36};\dfrac{-1}{-4}=\dfrac{1}{4}=\dfrac{9}{36}\)
mà 9<15
nên \(0< \dfrac{1}{4}< \dfrac{5}{12}\)
=>\(-1< -\dfrac{7}{9}< -\dfrac{1}{3}< 0< \dfrac{1}{4}< \dfrac{5}{12}\)
c: \(\dfrac{-1}{-2};0;\dfrac{3}{10};1;\dfrac{-2}{-5};\dfrac{3}{-4}\)
\(-\dfrac{3}{4}< 0\)
\(\dfrac{-1}{-2}=\dfrac{1}{2}=\dfrac{5}{10};\dfrac{3}{10}=\dfrac{3}{10};1=\dfrac{10}{10};\dfrac{-2}{-5}=\dfrac{4}{10}\)
mà 3<4<5<10
nên \(\dfrac{3}{10}< \dfrac{4}{10}< \dfrac{5}{10}< \dfrac{10}{10}\)
=>\(0< \dfrac{3}{10}< \dfrac{-2}{-5}< \dfrac{-1}{-2}< 1\)
=>\(-\dfrac{3}{4}< 0< \dfrac{3}{10}< \dfrac{-2}{-5}< \dfrac{-1}{-2}< 1\)
d: \(-\dfrac{37}{150}=\dfrac{-37}{150};\dfrac{17}{-50}=\dfrac{-17}{50}=\dfrac{-51}{150}\)
\(\dfrac{23}{-25}=\dfrac{-23}{25}=\dfrac{-138}{150};\dfrac{-7}{10}=\dfrac{-105}{150};\dfrac{-2}{5}=-\dfrac{60}{150}\)
mà -138<-105<-60<-51<-37
nên \(-\dfrac{138}{150}< -\dfrac{105}{150}< -\dfrac{60}{150}< -\dfrac{51}{150}< -\dfrac{37}{150}\)
=>\(\dfrac{23}{-25}< \dfrac{-7}{10}< \dfrac{-2}{5}< \dfrac{-17}{50}< \dfrac{37}{-150}\)
a: \(\dfrac{-7}{15}=\dfrac{-7\cdot4}{15\cdot4}=\dfrac{-28}{60}\)
\(\dfrac{5}{12}=\dfrac{5\cdot5}{12\cdot5}=\dfrac{25}{60}\)
b: \(\dfrac{1}{5}=\dfrac{1\cdot6}{5\cdot6}=\dfrac{6}{30}\)
\(\dfrac{-2}{3}=\dfrac{-2\cdot10}{3\cdot10}=\dfrac{-20}{30}\)
\(\dfrac{7}{10}=\dfrac{7\cdot3}{10\cdot3}=\dfrac{21}{30}\)
c: \(\dfrac{-15}{50}=\dfrac{-15\cdot3}{50\cdot3}=\dfrac{-45}{150}\)
\(\dfrac{9}{10}=\dfrac{9\cdot15}{10\cdot15}=\dfrac{135}{150}\)
\(\dfrac{26}{-30}=\dfrac{-26}{30}=\dfrac{-26\cdot5}{30\cdot5}=\dfrac{-130}{150}\)
d: \(\dfrac{7}{10}=\dfrac{7\cdot51}{10\cdot51}=\dfrac{357}{510}\)
\(\dfrac{-5}{-15}=\dfrac{1}{3}=\dfrac{1\cdot170}{3\cdot170}=\dfrac{170}{510}\)
\(\dfrac{3}{17}=\dfrac{3\cdot30}{17\cdot30}=\dfrac{90}{510}\)
e: \(\dfrac{-4}{-75}=\dfrac{4}{75}=\dfrac{4}{75}\)
\(\dfrac{-3}{5}=\dfrac{-3\cdot15}{5\cdot15}=\dfrac{-45}{75}\)
\(\dfrac{8}{25}=\dfrac{8\cdot3}{25\cdot3}=\dfrac{24}{75}\)
f: \(-\dfrac{4}{5}=\dfrac{-4\cdot7}{5\cdot7}=\dfrac{-28}{35}\)
\(\dfrac{6}{7}=\dfrac{6\cdot5}{7\cdot5}=\dfrac{30}{35}\)
Bài 2:
a: 2/5=14/35
3/7=15/35
b: -3/4=-9/12
-7/-12=7/12
c: 5/9=60/108
-11/-12=11/12=99/108
d: -4/7=-36/63
8/9=56/63
-10/21=-30/63
Bài 2:
a: 2/5=14/35
3/7=15/35
b: -3/4=-9/12
-7/-12=7/12
c: 5/9=60/108
-11/-12=11/12=99/108
d: -4/7=-36/63
8/9=56/63
-10/21=-30/63
a)
i.Ta có: BCNN(12, 30) = 60
60 : 12 = 5; 60 : 30 = 2. Do đó:
\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\) và \(\frac{7}{{30}} = \frac{{7.2}}{{30.2}} = \frac{{14}}{{60}}.\)
ii.Ta có: BCNN(2, 5, 8) = 40
40 : 2 = 20; 40 : 5 = 8; 40 : 8 = 5. Do đó:
\(\frac{1}{2} = \frac{{1.20}}{{2.20}} = \frac{{20}}{{40}}\)
\(\frac{3}{5} = \frac{{3.8}}{{5.8}} = \frac{{24}}{{40}}\)
\(\frac{5}{8} = \frac{{5.5}}{{8.5}} = \frac{{25}}{{40}}\).
b)
i.Ta có: BCNN(6, 8) = 24
24 : 6 = 4; 24: 8 = 3. Do đó
\(\begin{array}{l}\frac{1}{6} + \frac{5}{8} = \frac{{1.4}}{{6.4}} + \frac{{5.3}}{{8.3}}\\ = \frac{4}{{24}} + \frac{{15}}{{24}} = \frac{{19}}{{24}}.\end{array}\)
ii. Ta có: BCNN(24, 30) = 120
120: 24 = 5; 120: 30 = 4. Do đó:
\(\begin{array}{l}\frac{{11}}{{24}} - \frac{7}{{30}} = \frac{{11.5}}{{24.5}} - \frac{{7.4}}{{30.4}}\\ = \frac{{55}}{{120}} - \frac{{28}}{{120}} = \frac{{27}}{{120}} = \frac{9}{{40}}\end{array}\)
`#lv`
`-2/5+(-7/10)-9/6`
`=-2/5-7/10-9/6`
`=-4/10-7/10-9/6`
`=-11/10-3/2`
`=-11/10-15/10`
`=-26/10`
`=-13/5`
__
`1/2+(-2/5)-(-2/3)`
`=1/2-2/5+2/3`
`=5/10-4/10+2/3`
`=1/10+2/3`
`=3/30+20/30`
`=23/30`
__
`5/3-3/4+7/6`
`=20/12-9/12+14/12`
`=11/12+14/12`
`=25/12`
__
`-1/5+5/3-3/2`
`=-3/15+25/15-3/2`
`=22/15-3/2`
`=44/30-45/30`
`=-1/30`
__
`1/4-7/8+(-9/10)`
`=2/8-7/8-9/10`
`=-5/8-9/10`
`=-25/40-36/40`
`=-61/40`
__
`5/4+1/2+(-7/12)`
`=15/12+6/12-7/12`
`=21/12-7/12`
`=14/12`
`=7/6`
__
`-5/8-1/3+(-7/6)`
`=-5/8-1/3-7/6`
`=-15/24-8/24-28/24`
`=-52/24`
`=-13/6`
__
`-3/4-7/10+(-5/6)`
`=-3/4-7/10-5/6`
`=-45/60-42/60-50/60`
`=-137/60`
Bài làm
a, (-2/5) + (-7/10) - 9/6 =(-12/30) + (-21/30) - 45/30=(-33/30) - 45/30 = (-78/30)
b,1/2 + (-2/5) - (-2/3) = 15/30 + (-12/30) - (-20/30) = 3/30 - (-20/30) = 23/30
c,5/3 - 3/4 + 7/6 = 40/24 - 18/24 + 28/24 = 22/24 + 28/24 = 50/24 = 25/12
d, (-1/5) + 5/3 - 3/2 = (-6/30) + 50/30 - 45/30 = 44/30 - 45/30 = (-1/30)
e, 1/4 - 7/8 + (-9/10) = 10/40 - 35/40 - 36/40 = (-25/40) - 36/40 = (-61/40)
f, 5/4 +1/2 + (-7/12) = 15/12 + 6/12 - 7/12 = 21/12 - 7/12 = 14/12 = 7/6
g,(-5/8) -1/3 + (-7/6) = (-15/24) - 8/24 - 28/24 = (-23/24) - 28/24 = (-51/24)
k,(-3/4) - 7/10 + (-5/6) = (-45/60) - 42/60 - 50/60 = (-87/60) - 50/60 = (-137/60)
chúc bn học tốt nha
sai thì mn góp ý giúp mk
1-3+5-7+....+2009-2011+2013
=-2+(-2)+....+(-2)+2013
cÓ 503 SỐ HẠNG
=(-2).503 +2013
=1007
1. \(\frac{9}{30}=\frac{3}{10};\frac{98}{80}=\frac{49}{40};\frac{15}{1000}=\frac{3}{200}\)
Vì \(200⋮10;200⋮40\)
=> BCNN(10; 40; 200) = 200
200 : 10 = 20
200 : 40 = 5
=> \(\frac{3}{10}=\frac{3\cdot20}{10\cdot20}=\frac{60}{200}\), \(\frac{49}{40}=\frac{49\cdot5}{40\cdot5}=\frac{245}{200}\)
3: Số học sinh giỏi là 40*1/5=8 bạn
Số học sinh trung bình là 32*3/8=12 bạn
Số học sinh khá là 32-12=20 bạn
1:
a: -1/3+7/6=7/6-2/6=5/6
b: 5/7-3/5=25/35-21/35=4/35
c: 0,75*4/5=4/5*3/4=3/5
1: \(\dfrac{1}{5}=\dfrac{1\cdot10}{5\cdot10}=\dfrac{10}{50};\dfrac{-2}{50}=\dfrac{-2\cdot1}{50\cdot1}=\dfrac{-2}{50}\)
2: \(\dfrac{2}{-7}=\dfrac{-2}{7}=\dfrac{-2\cdot6}{7\cdot6}=\dfrac{-12}{42};\dfrac{-1}{-6}=\dfrac{-1\cdot\left(-7\right)}{-6\cdot\left(-7\right)}=\dfrac{7}{42}\)
3: \(\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot4}{3\cdot4}=\dfrac{-8}{12}\)
\(\dfrac{5}{-6}=\dfrac{-5}{6}=\dfrac{-5\cdot2}{6\cdot2}=-\dfrac{10}{12}\)
\(-\dfrac{7}{12}=\dfrac{-7\cdot1}{12\cdot1}=-\dfrac{7}{12}\)
1) Mẫu số chung: 50
\(\dfrac{1}{5}=\dfrac{1\cdot10}{5\cdot10}=\dfrac{10}{50}\)
Vậy ta được hai phân số đã quy đồng là \(\dfrac{10}{50}\) và \(\dfrac{-2}{50}\)
2) Mẫu số chung: 42
\(\dfrac{2}{-7}=\dfrac{2\cdot\left(-6\right)}{-7\cdot\left(-6\right)}=\dfrac{-12}{42}\)
\(\dfrac{-1}{-6}=\dfrac{-1\cdot\left(-7\right)}{-6\cdot\left(-7\right)}=\dfrac{7}{42}\)
Vậy ta được hai phân số đã quy đồng là \(\dfrac{-12}{42}\) và \(\dfrac{7}{42}\)
3) Mẫu số chung: 12
\(\dfrac{2}{-3}=\dfrac{2\cdot\left(-4\right)}{-3\cdot\left(-4\right)}=\dfrac{-8}{12}\)
\(\dfrac{5}{-6}=\dfrac{5\cdot\left(-2\right)}{-6\cdot\left(-2\right)}=\dfrac{-10}{12}\)
Vậy ta được hai phân số đã quy đồng là \(\dfrac{-8}{12}\); \(\dfrac{-10}{12}\) và \(\dfrac{-7}{12}\)