tinh ho to
A=2/1x2+2/2x3+2/3x4+......+2/99x101
B=(1+1/2)x(1+1/3)x(1+1/4)x.......x(1+1/2016)
C=3/1x4+3/4x7+3/7x10+............+3/64x67
D=(1-1/2)x(1-1/3)x(1-1/4)x............x(1-1/20)
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\(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{99x101}\)
\(A=2x\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{99x101}\right)\)
\(A=2x\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=2x\left(1-\frac{1}{101}\right)=2x\frac{100}{101}=\frac{200}{101}\)
------------------------------
\(B=\left(1+\frac{1}{2}\right)x\left(1+\frac{1}{3}\right)x\left(1+\frac{1}{4}\right)x...x\left(1+\frac{1}{2016}\right)\)
\(B=\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x...x\frac{2017}{2016}\) (rút gọn từ trên tử xuống dưới mẫu nhé)
\(B=\frac{2017}{2}\)
-------------------------------
\(C=\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+...+\frac{3}{64x67}\)
\(C=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{64}-\frac{1}{67}\)
\(C=1-\frac{1}{67}=\frac{67}{67}-\frac{1}{67}=\frac{66}{67}\)
--------------------------------
\(D=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x...x\left(1-\frac{1}{20}\right)\)
\(D=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{19}{20}\)(chỗ này cũng rút gọn từ trên tử xuống dưới mẫu)
\(D=\frac{1}{20}\)
Đề phải như này nhé
\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{99.100}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2.\left(1-\frac{1}{100}\right)\)
\(=2.\frac{99}{100}\)
Bài 1) Tính giá trị của M
M= 3 x 4 x 5 + 4 x 5 x 6 + ... + 13 x 14 x 15
Bài 2) Tính giá trị của A, biết rằng
M= 1/1x4 + 1/4x7 + 1/7x10+ ...+ 1/97x100
Bài 1:
$M=3.4.5+4.5.6+...+13.14.15$
$4M=3.4.5(6-2)+4.5.6(7-3)+....+13.14.15(16-12)$
$=-2.3.4.5+3.4.5.6-3.4.5.6+4.5.6.7+....-12.13.14.15+13.14.15.16$
$=-2.3.4.5+13.14.15.16=43560$
$M=43560:4=10890$
Bài 2:
a.
$3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}$
$=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{100-97}{97.100}$
$=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}$
$=1-\frac{1}{100}=\frac{99}{100}$
$M=\frac{99}{100}:3=\frac{33}{100}$
A=5/1x6+5/6x11+5/11x16+5/16x21+...+5/101x106
B=3/1x4+3/4x7+3/7x10+....+3/97x100
C=1/2x7+1/7x12+1/12x17+....+1/97x102
D=1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72
E=3/2x4+3/4x6+3/6x8+....+3/98x100
A = \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+\(\dfrac{5}{16.21}\)+...+\(\dfrac{5}{101.106}\)
A = \(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{106}\)
A = \(\dfrac{105}{106}\)
B = \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)
B = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
B = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
C = \(\dfrac{1}{2.7}+\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{97.102}\)
C= \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{97.102}\))
C = \(\dfrac{1}{5}\)\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\)+...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\) \(\dfrac{25}{51}\)
C = \(\dfrac{5}{51}\)
D = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)
D = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+ \(\dfrac{1}{8.9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{9}\)
D = \(\dfrac{8}{9}\)
E = \(\dfrac{3}{2.4}\)+\(\dfrac{3}{4.6}\)+\(\dfrac{3}{6.8}\)+...+\(\dfrac{3}{98.100}\)
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))
E = \(\dfrac{3}{2}\)\(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) \(\dfrac{49}{100}\)
E = \(\dfrac{147}{200}\)
1) Tính:
A= 1-2-3-4+5-6-7-8+9-10-11-12+...+97-98-99-100.
B= 1+2-3-4+5+6-7-8+9+10-11-12+...+197+198-199-200.
2) Tìm n € Z
a) n+2 chia hết n-1
b) n-7 chia hết 2n+3
c) n2-2 chia hết n + 3
d) n+5 chia hết 2n +1
3) Tìm x,y € N
a) 2xy + x - 4y = 10
b) 32y - 4xy - x= 12
c) ( x-2).(5y+1)=12
Bài 1: Tìm x:a) x - 2 + 7 = 1 . 3 . -9 b) 4 . x - 1 = (-5) 5 12 2 4 16 9 2 9c) -2 . x + 5 = 7 d) 2 . x : 7 . 1 = 29 3 8 12 5 3 8 15Bài 2: Tính:a) 2 + 1 . 4 b) 0,8 : [ 4 . 1,25] - [ 0,64 - 1 ] 5 7 5 25--------------------- 8 - 1 : 3 2 5 c) 8 : [ 7 - 12 ] d) ( 1,08 - 2 ) : 4 + ( 59 - 13 ). 36 9 2 7 25 7 9 4 17----------------------- [ 5 - 1 ] : 4
1:tính bằng cách thuận tiện nhất:
a)54 x 113 + 45 x 113 + 113 b)3/7 + 4/9 + 8/14 + 10/18
2:tính giá trị biểu thức
a)9/5 - 1/2 + 1/3
b)12/9 x 1/12 + 1/6 x 12/9
c)3/4 x (5/6 + 2/3)
d)3/2 - 2/3 : 2
1.a,=(54+45+1).113
=100.113
=11300
b,=(3/7+8/14)+(4/9+10/18)
=1+1
=2
2.a,=13/10+1/3
=49/30
b,=12/9.(1/12+1/6)
=12/9.1/4
=1/3
c,=3/4.3/2
=9/8
d,=3/2-1/3
=7/6
1:tính bằng cách thuận tiện nhất:
a)54 x 113 + 45 x 113 + 113
= 54 x 113 + 45 x 113 + 113x1
=113 x(54+45+1)
= 113x100
=1300
b)3/7 + 4/9 + 8/14 + 10/18
=(3/7+8/14)+(4/9+10/18)
= 1 + 1
=2
a) N = 1/2 + 1/6+ 1/12 + 1/20 + 1/30 +1/42 + 1/56 + 1/72 +1/90
b) 5/8 + 2/9 -2/5 + 3/8 + 4/9 + 1/3 - 3/5 =
c) 4 1/5 + 5 1/3 - 2 2/3 x 3 1/5 + 9/25 : 9/20 =
Bài 4: Tính các tổng sau:
a) 1 + 2 + 3 + 4 + ...... + n;
b) 2 +4 + 6 + 8 + .... + 2n;
c) 1 + 3 + 5 + ..... (2n + 1);
d) 1 + 4 + 7 + 10 + ...... + 2005;
e) 2 + 5 + 8 +......+ 2006;
g) 1 + 5 + 9 +....+ 2001.
a) \(1+2+3+4+...+n\)
\(=\left(n+1\right)\left[\left(n-1\right):1+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right):2\)
\(=n\left(n+1\right):2\)
\(=\dfrac{n\left(n+1\right)}{2}\)
b) \(2+4+6+..+2n\)
\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)
\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right)\)
\(=n\left(n+1\right)\)
c) \(1+3+5+...+\left(2n+1\right)\)
\(=\left[\left(2n+1\right)+1\right]\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}:2\)
\(=\left(2n+1+1\right)\left[\left(2n-1-1\right):2+1\right]:2\)
\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)
\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right)\)
\(=n\left(n+1\right)\)
d) \(1+4+7+10+...+2005\)
\(=\left(2005+1\right)\left[\left(2005-1\right):3+1\right]:2\)
\(=2006\cdot\left(2004:3+1\right):2\)
\(=2006\cdot\left(668+1\right):2\)
\(=1003\cdot669\)
\(=671007\)
e) \(2+5+8+...+2006\)
\(=\left(2006+2\right)\left[\left(2006-2\right):3+1\right]:2\)
\(=2008\cdot\left(2004:3+1\right):2\)
\(=1004\cdot\left(668+1\right)\)
\(=1004\cdot669\)
\(=671676\)
g) \(1+5+9+...+2001\)
\(=\left(2001+1\right)\left[\left(2001-1\right):4+1\right]:2\)
\(=2002\cdot\left(2000:4+1\right):2\)
\(=1001\cdot\left(500+1\right)\)
\(=1001\cdot501\)
\(=501501\)
CMR
a)A= 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/n(n+3) < 1
b)2/5<B=1/2 mũ 2 +1/3 mũ 2 + 1/4 mũ 2 + . . . + 1/9 mũ 2 +1/10 mũ 2
c)C=1/4 mũ 2 + 1/6 mũ 2 + 1/8 mũ 2 + . . . + 1/(2n) mũ 2 <1/4
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