x^2 - x - 2004 + 2005
I, Tìm x: a, \(\dfrac{x-2004}{2003}+\dfrac{x-2003}{2005}+\dfrac{x-2005}{2004}=3+\dfrac{2005}{2004}+\dfrac{2004}{2005}\)
Tính nhanh :
a ) 2 x 3 x 4 x 5 x 7 x 8 x 25 x 125
b ) 19001570 x ( 20052005 x 2004 - 20042004 x 2005 )
c ) 2004 x 2004 + 3006
2005 x 2005 - 1003
Làm 1 câu tặng 3 k
2 câu 6 k
3 câu 9 k
Nha
Please
a, 2 x 3 x 4 x 5 x 7 x 8 x 25 x 125 = 10 x 100 x 1000 x 3 x 7 = 21000000
Tìm x: a, \(\frac{x-2004}{2003}+\frac{x-2003}{2004}+\frac{x-2005}{2004}=3+\frac{2005}{2003}\)\(+\frac{2004}{2005}\)
c) 22/5 + 51/9 + 11/4 + 3/5 + 1/3 + 1/4
= 22/5 +3/5 +51/9 + 1/3 +11/4+1/4
= (22/5 +3/5) +(51/9 + 3/9) +(11/4+1/4)
= 25/5 +54/9 +12/4
= 5 +6 +3
= 14
d) (1/6 + 1/10 + 1/15) : (1/6 + 1/10 - 1/15)
= (5/30 + 3/30 +2/30 ) :(5/30 +3/30 -2/30)
= 10/30 : 6/30
= 1/3 : 1/5
= 5/3
2003 x 2004 + 2005 x15 + 1989
2004 x 2005 - 2004 x 2002
Mình làm cau thứ hai còn câu trước giống như vậy
2004 x 2005 - 2004 x 2002
=2004 x (2005 -2002)
=2004 x 3
=6012tick nha
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
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
Giải phương trình sau :
\(\frac{x^2-2008}{2007}+\:\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\:\frac{x^2-\:2005}{2004}+\:\frac{x^2-2004}{2003}+\:\frac{x^2-2003}{2002}\)
Ta có : \(\frac{x^2-2008}{2007}+\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\frac{x^2-2005}{2004}+\frac{x^2-2004}{2003}+\frac{x^2-2003}{2002}\)
=> \(\frac{x^2-2008}{2007}+1+\frac{x^2-2007}{2006}+1+\frac{x^2-2006}{2005}+1=\frac{x^2-2005}{2004}+1+\frac{x^2-2004}{2003}+1+\frac{x^2-2003}{2002}+1\)
=> \(\frac{x^2-2008}{2007}+\frac{2007}{2007}+\frac{x^2-2007}{2006}+\frac{2006}{2006}+\frac{x^2-2006}{2005}+\frac{2005}{2005}=\frac{x^2-2005}{2004}+\frac{2004}{2004}+\frac{x^2-2004}{2003}+\frac{2003}{2003}+\frac{x^2-2003}{2002}+\frac{2002}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}=\frac{x^2-1}{2004}+\frac{x^2-1}{2003}+\frac{x^2-1}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}-\frac{x^2-1}{2004}-\frac{x^2-1}{2003}-\frac{x^2-1}{2002}=0\)
=> \(\left(x^2-1\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
=> \(x^2-1=0\)
=> \(x^2=1\)
=> \(x=\pm1\)
Vậy phương trình có 2 nghiệm là x = 1, x = -1 .
So sánh A và B biết:
A = 2003 x 2004 - 1/2003 x 2004
B = 2004 x 2005 - 1/2004 x 2005
Tính nhanh:
a)1153 + 1153 + 1004 - 513 - 513 - 1000
b)(2003 x 2004 + 2004 x 2005 ) x (2005 : 1 - 1 x 2005 )
2005 x 2004 -1 / 2003 x 2005 + 2004