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- ( 1 + (-2) ) + ( 3 + (-4) ) + . . . + (19 + (-20))
= -1 + -1 +....+-1
= -10 - (1 – 2) + (3 – 4) + . . . + (99 – 100)
= -1 + -1 +...+-1
= -50 - (2 – 4) + (6 – 8) + . . . + (48 – 50)
= -2 + -2 +...+-2
= -50
1. 1+-(2)+3+(-4)+......+19+(-20)=(-1)+(-1)+....+(-1)=(-1).10=-10
2.1-2+3-4+.....+99-100=-1+-1+...+-1=-1.50=-50
1.
Ta có:
1/2 < 2/3
3/4 < 4/5
.............
99/100 < 100/101
=> 1/2*3/4*5/6*...*99/100 < 2/3*4/5*6/7*...*100/101
=> A < B
2.
\(A\cdot B=\left[\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right]\cdot\left[\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right]\)
\(A\cdot B=\frac{\left[1\cdot3\cdot5\cdot7\cdot...\cdot99\right]\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]}{\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]\left[3\cdot5\cdot7\cdot9\cdot...\cdot101\right]}=\frac{1\cdot3\cdot5\cdot...\cdot99}{3\cdot5\cdot7\cdot...\cdot101}=\frac{1}{101}\)
3.
Vì A < B => A.A < A.B => A2 < 1/101 < 1/100
Mà A2 < 1/100 <=> A2 < \(\frac{1}{10}^2\)=> A < 1/10
\(1,\)
\(A=-\frac{7}{12}+\frac{12}{18}+\frac{5}{4}\)
\(=-\frac{7}{12}+\frac{2}{3}+\frac{5}{4}\)
\(=-\frac{7}{12}+\frac{8}{12}+\frac{15}{12}\)
\(=\frac{-7+8+15}{12}\)
\(=\frac{4}{3}\)
\(1,\)
\(B=\frac{1}{4}-\frac{8}{7}:8-3:\frac{3}{4}.\left(-2\right)^2\)
\(=\frac{1}{4}-\frac{8}{7}.\frac{1}{8}-3.\frac{4}{3}.4\)
\(=\frac{1}{4}-\frac{1}{7}-16\)
\(=\frac{7-4-448}{28}\)
\(=-\frac{445}{28}\)
Mình đang bận, bạn cần gấp thế à? Trả lời trong tin nhắn nhé!!!
1. (-99).98.(-97)>0
2. (-5).(-4).(-3).(-2).(-1)<0
3. (-245).(-47).(-199)<123.(+315)
4. 2987.(-1974).(+243).0=0
5. (-12).(-45):(-27)<|-1|
tích nha
- (-99) . 98 . (-97) < 0
- (-5) . (-4) . (-3) . (-2) . (-1) < 0
- (-245) . (-47) . (-199) < 123 . (+315)
- 2987 . (-1974) . (+243) . 0 = 0
- (-12) . (-45) : (-27) với <│-1|
1)C= 1/5+1/10+1/20+1/40+...+1/1280
\(=\frac{1}{5}\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
Đặt cái trong ngoặc là A ta có:\(2A=2+1+...+\frac{1}{2^7}\)
\(2A-A=\left(2+1+...+\frac{1}{2^7}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)
\(A=2-\frac{1}{2^8}\).Thay A vào ta được:\(C=\frac{1}{5}\left(2-\frac{1}{2^8}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)
2)D= 2/1*3+2/3*5+2/5*10+2/7*9+2/9*11+2/11*18+2/13*15
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
\(=\frac{14}{15}\)
3)E= 4/3*7+4/7*11+4/11*15+4/15*19+4/19*23+4/23*27
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}\)
\(=\frac{8}{27}\)
4)G= 1/2+1/6+1/12+1/20+1/30+1/42+...+1/110
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
5)H= 3/1*2+3/2*3+3/3*4+3/4*5+...+3/9*10
\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=3\left(1-\frac{1}{10}\right)\)
\(=3\times\frac{9}{10}\)
\(=\frac{27}{10}\).Lần sau bạn đăng ít một thôi nhé