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\(a,1-2+3-4+5-6+......+199-200\)
\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+.....+\left(199-200\right)\)( 100 cặp )
\(=-1+\left(-1\right)+\left(-1\right)+........+\left(-1\right)\)( 100 số hạng )
\(=-1.100\)
\(=-100\)
\(a.1-2+3-4+5-6+...+199-200\)
\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(199-200\right)\) (có tất cả \(200:2=100\)cặp)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(=\left(-1\right).200=-200\)
\(b.1+2-3-4+5+6-7-8+...+97+98-99-100\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\) (có \(100:4=25\)cặp)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(=\left(-4\right).25=-100\)
\(c.1+\left(-6\right)+11+\left(-16\right)+...+21+\left(-26\right)\)
\(=\left[1+\left(-6\right)\right]+\left[11+\left(-16\right)\right]+...+\left[21+\left(-26\right)\right]\) (có tất cả \(26:2=13\)cặp)
\(=\left(-5\right)+\left(-5\right)+...+\left(-5\right)\)
\(=-5.13=-65\)
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b)\(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\)
\(=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{61}-\frac{1}{66}\)
\(=\frac{1}{11}-\frac{1}{66}\)
\(=\frac{5}{66}\)
a,\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
ta có:
\(\frac{1}{1.2}=\frac{2-1}{1.2}=\frac{2}{1.2}-\frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{2.3}=\frac{3-2}{2.3}=\frac{3}{2.3}-\frac{2}{2.3}=\frac{1}{2}-\frac{1}{3}\)
...
\(\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}=\frac{99}{100}\)
b,
\(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.16}+...+\frac{5}{61.66}\)
ta có:
\(\frac{5}{11.16}=\frac{16-11}{11.16}=\frac{16}{11.16}-\frac{11}{11.16}=\frac{1}{11}-\frac{1}{16}\)
\(\frac{5}{16.21}=\frac{21-16}{16.21}=\frac{21}{16.21}-\frac{16}{16.21}=\frac{1}{16}-\frac{1}{21}\)
...
\(\frac{5}{61.66}=\frac{66-61}{61.66}=\frac{66}{61.66}-\frac{61}{61.66}=\frac{1}{61}-\frac{1}{66}\)
= \(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)
=\(\frac{1}{11}-\frac{1}{66}\)=\(\frac{5}{66}\)
Bài 1:
a) \(\frac{16}{15}.\frac{\left(-5\right)}{14}.\frac{54}{24}.\frac{56}{21}\)
\(=\frac{4.2.2}{5.3}.\frac{\left(-5\right)}{2.7}.\frac{3.3}{4}.\frac{8}{3}\)
\(=\frac{4.2.2.\left(-5\right).3.3.8}{5.3.2.7.4.3}\)
\(=\frac{-16}{7}\)
b) \(\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{15}{21}.\frac{4}{\left(-5\right)}\)
\(=\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{5}{7}.\frac{2.2}{\left(-5\right)}\)
\(=\frac{7.\left(-5\right).5.2.2}{3.2.7.\left(-5\right)}\)
\(=\frac{10}{3}\)
Bài 2:
a) \(\frac{21}{24}.\frac{11}{9}.\frac{5}{7}=\frac{7}{8}.\frac{11}{9}.\frac{5}{7}=\frac{11.5}{8.9}=\frac{55}{72}\)
b) \(\frac{5}{23}.\frac{17}{26}+\frac{5}{23}.\frac{9}{26}\)
\(=\frac{5}{23}.\left(\frac{17}{26}+\frac{9}{26}\right)=\frac{5}{23}.1=\frac{5}{23}\)
c) \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}=\frac{3}{29}.\frac{29}{3}-\frac{1}{5}.\frac{29}{3}\)
\(=1-1\frac{14}{15}=\frac{14}{15}\)
Bài 3:
a) x/5 = 2/5
=> x =2
b) -4/x = 20/14 = 10/7
=> -4/x = 10/7
=> x.10 = (-4).7
x.10 = - 28
x= -28 :10
x= -2,8
c) 4/7 = 12/x = 12/ 21
=> 12/x = 12/21
=> x = 21
d) 3/7 = x / 21 = 9/21
=> x/21 = 9/21
=> x= 9
`a)1 4/23 + ( 5/21-4/23)+16/21-1/2`
`=27/23+5/21-4/23+16/21-1/2`
`=(27/23-4/23)+(5/21+16/21)-1/2`
`=23/23+21/21-1/2`
`=1+1-1/2`
`=2-1/2`
`=4/2-1/2`
`=3/2`
___
`b)75%-(5/2+5/3)+(-1/2)^3`
`=3/4-5/2+5/3+(-1/8)`
`=(3/4-5/2-1/8)+5/3`
`=(6/8-20/8-1/8)+5/3`
`=-15/8+5/3`
`=-45/24+40/24`
`=-5/24`
___
`c)-3/4(-55/9).8/11`
`=-3/4.(-40/9)`
`=-10/3`
__
`d)-3/8 . 6/13 + 7/13 . (-3/8) + 1 3/8`
`= -3/8 . (6/13 + 7/13) + 11/8`
`= -3/8 . 13/13 + 11/8`
`= -3/8 .1 + 11/8`
`= -3/8 + 11/8`
`= 8/8`
`=1`
a. \(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)
\(=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)
b. \(D=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{97.100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
\(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-....-\frac{1}{66}\)
\(C=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)
\(D=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-....-\frac{1}{100}\right)\)
\(D=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)