a)[3.(-7)7-10.(-7)7] :(-7)6
b)(-11)9:[5.(-11)5-16.(-11)5]
c)(-1).(-1)2.(-1)3...(-1)100
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C=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{100}-\left(\frac{1}{2.1}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\frac{99}{100}\)
=\(\frac{-98}{100}=\frac{-49}{50}\)
C=1/100 -1/100.99 -1/99.98 -1/98.97-......- 1/3.2 -1/2.1
= 1/100 - (1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1)
Đặt A = 1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1 => C = 1/100 - A
Dễ thấy 1/2.1 = 1/1 - 1/2
1/3.2 = 1/2 - 1/3
.....................
1/99.98 = 1/98 - 1/99
1/100.99 = 1/99 - 1/100
=> cộng từng vế với vế ta
A = 0-1 + 2-3 + 4-5 +...+ 2017-2018
=> A = (-1) + (-1) + (-1) +...+ (-1) (Có 1009 số hạng)
=> A = 1009.(-1)
=> A = -1009
B = 1-3+5-7+ 9-11+....+2005-2007
=> B = (-2) + (-2) +(-2) +...+ (-2) (Có 502 số hạng)
=> B = 502.(-2)
=> B = -1004
C=1+2+3-4-5-6+7+8+9-10-11-12+.....+97+98+99-100-101-102
=> C = (1+2+3-4-5-6)+...+(97+98+99-100-101-102) (có 17 cặp số)
=> C = (-9) + (-9) +...+ (-9) (có 17 số hạng)
=> C = (-9).17
=> C = -153
1:
a: =8/7-5/88=669/616
b: \(=1+\dfrac{2}{9}\cdot\dfrac{3}{7}-\dfrac{10}{7}=1+\dfrac{2}{21}-\dfrac{10}{7}\)
\(=\dfrac{21+2-30}{21}=\dfrac{-7}{21}=\dfrac{-1}{3}\)
c: \(=\dfrac{11}{3}-\dfrac{6}{7}+4=\dfrac{77-18+84}{21}=\dfrac{143}{21}\)
Bài 2:
a: =>9/4-x=5/11*2=10/11
=>x=9/4-10/11=59/44
b: =>2/9:x=19/21
=>x=2/9:19/21=14/57
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
B=4*13/9*3-4/3*40/9
B=4/3*13/9-4/3*40/9
B=4/3*(13/9-40/9)
B=4/3*(-27)/9
B=4*(-3)/9
B=-4
A=6/7 + 1/7.(2/7+5/7)
A=6/7 + 1/7.7/7=6/7+1/7.1
A=6/7+1/7=7/7=1
a: \(=\dfrac{3\cdot\left(-7\right)^7-10\cdot\left(-7\right)^7}{\left(-7\right)^6}=3\cdot\left(-7\right)-10\cdot\left(-7\right)=-21+70=49\)
b: \(=\dfrac{\left(-11\right)^9}{\left(-11\right)^5\cdot\left(-11\right)}=\left(-11\right)^3=-1331\)