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\(E=\frac{2}{10x12}+\frac{2}{12x14}+...+\frac{2}{108x110}\)
\(\Rightarrow E=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{108}-\frac{1}{110}\)
\(\Rightarrow E=\frac{1}{10}-\frac{1}{110}=\frac{10}{110}=\frac{1}{11}\)
Công thức :\(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)
\(E=\frac{2}{10\cdot12}+\frac{2}{12\cdot14}+\frac{2}{14\cdot16}+...+\frac{2}{108\cdot110}\)
\(E=\frac{1}{10\cdot12}+\frac{1}{12\cdot14}+\frac{1}{14\cdot16}+...+\frac{1}{108\cdot110}\)
\(E=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{108}-\frac{1}{110}\)
\(E=\frac{1}{10}-\frac{1}{110}\)
\(E=\frac{11}{110}-\frac{1}{110}\)
\(E=\frac{10}{110}\)\(=\frac{1}{11}\)
Học tốt nha bạn !!!
(1/10+-1/10)+(1/11+-1/11)+(1/12+-1/12)+(-1/13+1/13)+(-1/14+1/14)+(-1/15+1/15)+1/16
=0 + 0 +0 + 0 +0 +0 +1/16
=1/16
1/2+1/4+1/6+1/8+1/10+1/12+1/14+1/16+1/18+1/20=(1/2+1/8)+(1/4+1/6)+(1/10+1/20)+(2/12+1/18)+(1/14+1/16)
=5/8+5/12+3/20+2/9+15/112
=25/24+67/180+15/112
=509/360+15/112
=143/120
a,
\(2-4+6-8+...+1998-2000\)( có 1000 số )
\(=\left(2-4\right)+\left(6-8\right)+.....+\left(1998-2000\right)\)( có 500 nhóm )
\(=\left(-2\right)+\left(-2\right)+....+\left(-2\right)\)( có 500 số -2 )
\(=-2.500\)
\(=-1000\)
b.
\(\text{11-12+13-14+15-16+17-18+19-20}\)( có 10 số hạng )
\(=\left(11-12\right)+\left(13-14\right)+...+\left(19-20\right)\)( có 5 nhóm )
\(=-1+\left(-1\right)+....+\left(-1\right)\)
\(=-1.5\)
\(=-5\)
c,
\(\text{101-102-(-103)-104-(-105)-106-(-107)-108-(-109)-110}\)
\(=101-102+103-104+....+109-110\)( có 10 số )
\(=-1+\left(-1\right)+...+\left(-1\right)\)( có 5 nhóm )
\(=-1.5\)
\(=-5\)
1+2+3+...+99+100
=(1+100)x50
=101x50
=5050
2+4+6+8+10+...+18
=(2+18):2 x9
=90
1+2+3+4+...+200
=(1+200)x 100
=201 x100
=20100
1+2+3+...+99+100
=(1+100)x50
=101x50
=5050
2+4+6+8+10+...+18
=(2+18):2 x9
=90
1+2+3+4+...+200
=(1+200)x 100
=201 x100
=20100
1+(2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17)x0,000001
=1+(2+(3+17x10)+10)x0,000001
=1+(2+200+10)x0,000001
=1+212x0,000001
=1+0,000212
=1,000212
1+2-3-4+5+6-...+301+302
đặt A=(1+2)+(5+6)+...+(301+302)=3+11+19+...+603
A=[(603-3):8+1]x(603+3)=75x606=45450
đặt B=-3-4-7-8-11-13-...-299-300
B=(-3-4)+(-7-8)+(-11-12)+..+(-299-300)
B=-(7+15+23+...+599)
B=-[(599-7):8+1]x(599+7)=45450
ta có:
1+2-3-4+5+6-7-8+...+301+302=A-B=45450-45450=0
a)\(\frac{4}{5}-\frac{1}{4}+\frac{3}{10}\)
\(=\frac{16}{20}-\frac{5}{20}+\frac{6}{20}\)
\(=\frac{17}{20}\)
b) \(\frac{2}{5}:\left(1-\frac{1}{10}\right)\)
\(=\frac{2}{5}:\frac{9}{10}\)
\(=\frac{4}{9}\)
c)\(\frac{7}{8}\times\frac{4}{9}+\frac{1}{14}:\frac{5}{14}\)
\(=\frac{7}{18}+\frac{1}{5}\)
\(=\frac{53}{90}\)
d)\(\frac{2}{7}\times\frac{3}{11}+\frac{2}{7}\times\frac{8}{11}\)
\(=\frac{2}{7}\times\left(\frac{3}{11}+\frac{8}{11}\right)\)
\(=\frac{2}{7}\times1=\frac{2}{7}\)
e) \(12+\left(16-11\right)\times4\)
\(=12+20=32\)
f)\(2\frac{3}{7}+1\frac{4}{7}\)
\(=\frac{17}{7}+\frac{11}{7}\)
\(=4\)
g)\(\frac{2}{3}\times\frac{4}{5}+\frac{1}{5}:\frac{9}{11}\)
\(=\frac{8}{15}+\frac{11}{45}\)
\(=\frac{7}{9}\)
h)\(\left(6,2:2+3,7\right):0,2\)
\(=\left(3,1+3,7\right):0,2\)
\(=6,8:0,2=34\)
#H
2F = 2/10 x 12 + 2 / 12 x 14 + 2 / 14 x 16 + ... + 2 / 108 x 110
= 1/10 - 1/12 + 1/12 - 1/14 + 1/14 - 1/16 + ... + 1/108 -1/110
= 1/10 - 1/110
= 1/11