tính nhanh
a) 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + ......... + 1/98*99 + 1/99*100
b) 5/11*16 + 5/16*21 + 5/21*16 + .......... + 5/61*66
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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. \(D=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{97.100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
\(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-....-\frac{1}{66}\)
\(C=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)
\(D=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-....-\frac{1}{100}\right)\)
\(D=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
\(a,1-2+3-4+5-6+......+199-200\)
\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+.....+\left(199-200\right)\)( 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)
\(=\left(-5\right)+\left(-5\right)+...+\left(-5\right)\)
\(=-5.13=-65\)
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
1. Tính nhanh:
a, 1/1 * 2 + 1/2 * 3 + 1/3 * 4
b, 5/11 *16 + 5/16 * 21 + 5/21 * 26
c, 1/20 + 1/30 + 1/42
a) 1/1 x 2 + 1/2 x 3 + 1/3 X 4 = 29/6
b) 5/11 x 16 + 5/16 x 21 + 5/21 x 26 = 74015/3696
c) 1/20 +1/30 + 1/42 = 3/28
nho k minh nha
a) 1 + 1/4 + 1/8 + 1/16
= 16/16 + 4/16 + 2/16 + 1/16
= 23/16
b) 2 - 1/8 - 1/12 - 1/16
= 96/48 - 6/46 - 4/48 - 3/48
= 83/48
c) 4/99 × 18/5 : 12/11 + 3/5
= 8/55 : 12/11 + 3/5
= 2/15 + 3/5
= 2/15 + 9/15
= 11/15
d) (1 - 3/4) × (1 + 1/3) : (1 - 1/3)
= 1/4 × 4/3 : 2/3
= 1/3 : 2/3
= 2
A = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 +....+ 99 - 100
A = (1 - 2) + ( 3- 4) + ....+ (99 - 100)
Xét dãy số 1; 3;...; 99
Dãy số trên là dãy số cách đều với khoảng cách là: 3 - 1 = 2
Số số hạng của dãy số trên là: ( 99 - 1): 2 + 1 = 50
A là tổng của 50 nhóm mỗi nhóm cóa giá tri là: 1 - 2 = - 1
A = - 1 \(\times\) 50 = - 50
B = 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 +...+ 97 - 98 - 99 + 100
B = ( 1 - 2 - 3 + 4) + ( 5 - 6 - 7 + 8) +...+ ( 97 - 98 - 99 + 100)
B = 0 + 0 +...+ 0
B = 0
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:
\(\frac{5}{11.16}=\frac{16-11}{11.16}=\frac{16}{11.16}-\frac{11}{11.16}=\frac{1}{11}-\frac{1}{16}\)
\(\frac{5}{16.21}=\frac{21-16}{16.21}=\frac{21}{16.21}-\frac{16}{16.21}=\frac{1}{16}-\frac{1}{21}\)
...
\(\frac{5}{61.66}=\frac{66-61}{61.66}=\frac{66}{61.66}-\frac{61}{61.66}=\frac{1}{61}-\frac{1}{66}\)
= \(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)
=\(\frac{1}{11}-\frac{1}{66}\)=\(\frac{5}{66}\)