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c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)
\(=1-2=-1\)
Giải:
a)-1/12+4/3=-1/12+16/12=15/12=5/4
b)(-4/14-3/15)-(1/5-20/35-(-1)).7
=-17/35-22/35.7
=-17/35-22/5
=-171/35
c)3/5+-5/20+30/75+-7/4
=3/5+-1/4+2/5+-7/4
=(3/5+2/5)+(-1/4+-7/4)
=1+-2
=-1
d)5/6.-12/14+7/13
=-5/7+7/13
=-16/91
e)2/-9-5/-36-1/4
=-1/12-1/4
=-1/3
f)2/23+-5/12+7/18+21/23+-7/12
=(2/23+21/23)+(-5/12+-7/12)+7/18
=1+-1+7/18
=7/18
1: =72/90+65/90=137/90
2: =24/56-77/56=-53/56
3: =-7/10+4/5=1/10
4: =15/100-4/100=11/100
5: =4/6-5/6=-1/6
6: =10/40-15/40-76/40=-81/40
7: =-9/10+7/18
=-81/90+35/90=-46/90=-23/45
8: =27/90-55/90=-28/90=-14/45
9: =36/60-50/60-35/60=-49/60
10: =-4/9+5/6-3/8
=-32/72+60/72-27/72
=1/72
1+2+3+4+5+6+7+8+9+10=55
11+12+13+14+15+16+17+18+19+20=155
1+2+3+4+5+6+7+8+9+10+11+12+13+14 +15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30-50-53=362
câu a :
\(\dfrac{-8}{24}+\dfrac{-4}{12}=\dfrac{-1}{3}+\dfrac{-1}{3}=\dfrac{-2}{3}\)
câu b :
\(\dfrac{-20}{35}+\dfrac{16}{24}=\dfrac{-4}{7}+\dfrac{2}{3}=\dfrac{2}{21}\)
câu c :
\(\dfrac{-3}{9}+\dfrac{-6}{15}=\dfrac{-1}{3}+\dfrac{-2}{5}=\dfrac{-11}{15}\)
câu d :
\(\dfrac{3}{13}-\dfrac{4}{10}=\dfrac{3}{13}-\dfrac{2}{5}=\dfrac{-11}{65}\)
câu e :
\(\dfrac{5}{17}-\dfrac{9}{15}=\dfrac{5}{17}-\dfrac{3}{5}=\dfrac{-26}{85}\)
câu g :
\(\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{3}{-9}=\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{-3}{9}\\ =\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{-1}{3}=\dfrac{-7}{30}\)
câu h :
\(\dfrac{5}{4}-\dfrac{1}{2}+\dfrac{-7}{8}=\dfrac{10}{8}-\dfrac{4}{8}+\dfrac{-7}{8}=\dfrac{-1}{8}\)
a; A = -1 + 3 - 5 + 7 - 9 + 11 - 13 + 15 - 17
A = (-1 + 11) + ( 3 - 13) + (-5 + 15) + (7 - 17) - 9
A = 10 - 10 + 10 - 10 - 9
A = (10 - 10) + (10 - 10) - 9
A = 0 + 0 - 9
A = -9
b; B = 1+2-3-4+5+6-7-8+9+10-11-12+13+14-15-16+17+18-19-20
B = (1+2-3-4) + (5+6-7-8)+(9+10-11-12)+(13+14-15-16)+(17+18-19-20)
B= -4+(-4)+(-4)+(-4)+(-4)
B= -4 . 5
B= -20
\(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{870}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{29.30}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{29}-\frac{1}{30}\)
\(=1-\frac{1}{30}\)
\(=\frac{29}{30}\)