2004x2005+2006x6-6/ 2005 x 1997 +4 x2005
2004x2005+2006x6-6/2005x1997+4x2005
\(\dfrac{2004\times2005+2006\times6-6}{2005\times1997+4\times2005}\\ =\dfrac{2004\times2005+\left(2005+1\right)\times6-6}{2005\times\left(1997+4\right)}\\ =\dfrac{2004\times2005+2005\times6+6-6}{2005\times2001}\\ =\dfrac{2005\times\left(2004+6\right)}{2005\times2001}\\ =\dfrac{2010}{2001}\\ =\dfrac{670}{667}\)
Tính nhanh:
a.2,3+3,4+4,5+...+11,1+12,2
b.2004x2005+2006x6-6 / 2005x197+4x2005
Tính kết quả sau:
[ 2003 x 2004 + 2004 x 2005} x{ 2005:1 - 1 x2005}
[ 2003 x 2004 + 2004 x 2005} x { 2005 : 1 - 1 x 2005}
= 8032032 x 0 = 0
Tính kết quả sau: { 2003 x 2004+ 2004 x 2005 }x { 2005 :1 -1 x2005}
={2003 x 2004 x 2005} x {2005 - 2005}
={2003 x 2004 x 2005} x 0
=0
={2003 x 2004 x 2005} x {2005 - 2005}
={2003 x 2004 x 2005} x 0
=0
2004x2005+2006x6-6
2005x1997+4x2005
\(\frac{670}{667}\)
A.\(\frac{2004x2005+2006x6-6}{2005x1997+4x2005}\) \(\frac{1999x2000+2001x5-5}{504x2000+500x2000}\)
B.\(\frac{2003x4+1998+2001x2002}{2002+2002x502+500x2002}\) \(\frac{2000x4+1995+2001x1995}{1995x495+1995x5+1995x3}\)
C.\(\frac{72:2x574+286x2x64}{4+4+8+12+20+...+220}\) \(\frac{72+36x2+24x3+18x4+12x6+168}{2+2+4+6+...+512+1024}\)
1
tìm các sồ tự nhiên n sao cho n+6chia het cho n-4
2
tìm các số nguyên x1,x2,x3,....,x2005 thỏa mãn x1+x2+x3+....,+x2005=0 va x1+x2=x3+x4=x2005+1=1
1, n+ 6\(⋮\)n-4
n-4\(⋮\)n-4
=> (n+6)-(n-4)\(⋮\)n-4
n+6-n+4\(⋮\)n-4
( n gạch hết )
=> 6+4\(⋮\)n-4
10\(⋮\)n-4
=> n-4\(\in\)ước của 10
Ư(10)={ -10; -5;-2;-1;1;2;5;10}
n-4 | n |
-10 | -14 |
-5 | -9 |
-2 | -6 |
-1 | -5 |
1 | 5 |
2 | 6 |
5 | ;9 |
10 | 14 |
Vậy x=-14;-9;-6;-5;5;6;9;14
Tính bằng cách hợp lý
a, 5 - ( 1997 - 2005 ) +1997
b, 4567 +(1234 -4567 ) - 4
c, 173 - ( 36 + 173 ) + [( 175 - 27) -175 - (-36 +50 )]
d, 3 x ( 25 : 5 -14 : 2 ) -5 x ( 6 : 2 )
dễ mà : a, 5 - ( 1997 - 2005 ) +1997
= ( 2005 + 5 ) - ( 1997 - 1997 )
= 2010 - 0
= 2010
a, 5 - (1997 -2005 ) + 1997
= 5 - 1997 + 2005 +1997
= (5 + 2005 ) - ( 1997 - 1997)
= 2010 - 0
= 2010
b, 4567 + (1234 - 4567 )-4
= 4567 + 1234 - 4567 -4
= (4567 - 4567 ) + (1234 - 4)
= 0 + 1230
= 1230
c, 173 - ( 36 + 173 ) + [(175 - 27 ) - 175 - ( -36 + 50 )]
= 173 - 36 - 173 + 175 - 27 - 175 - 36 - 50
= ( 173 - 173 ) + (175 - 175 ) - ( 36 + 36 ) - (27 + 50)
= 72 - 77
= -5
d, 3 . (25 : 5 - 14 : 2 ) - 5 . ( 6 : 2)
= 3 (-2) - 5 .3
= 3 ( -2 - 5 )
= 3 . (-7 )
= - 21
a) 5-(1997-2005)+1997
= 5-(-1997)+2005+1997
= (5+2005)+(-1997+1997)
= 2010+0
= 0
b) 4567+(1234-4567)-4
làm tương tự câu a
Tính giá trị biểu thức sau
1, A=6+5^2+5^3+5^4 +....+5^1996+5^1997
2,B=10+9^2+9^3+9^4+....+9^2004+9^2005
3, C=x^20-2006x^19+x^2018-2006x^17+....+2006x^2-2006x+2006 với x=2005