2+(-4)+6+(-8)+...+x=2000
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18 tháng 6 2019
(X -10/1994 -1) + (X-8/1996 - 1) + (X-6/1998 - 1)+ (X-4/2000 - 1) + (X-2/2002 - 1) = (X-2002/2 - 1) + (X-2000/4 - 1) + (X-1998/6 - 1) + (X-1996/8 - 1) + (X-1994/10 - 1)
=> x-2004/1994 + x-2004/1996 + x-2004/1998 + x-2004/2000 + x-2004/2002 = x-2004/2 + x-2004/4 + x-2004/6 + x-2004/8 + x-2004/1994
=> x-2004/1994 + x-2004/1996 + x-2004/1998 + x-2004/2000 + x-2004/2002 - x-2004/2 - x-2004/4 - x-2004/6 - x-2004/8 - x-2004/1994 = 0
=> (x - 2004)(1/994 + 1/1996 + 1/1998 + 1/2000 + 1/2002 + 1/2 + 1/4 + 1/6 + 1/8) = 0
Mà (1/994 + 1/1996 + 1/1998 + 1/2000 + 1/2002 + 1/2 + 1/4 + 1/6 + 1/8) \(\ne\)0
=> x - 2004 = 0
=> x = 2004
Vậy x = 2004
LT
25 tháng 2 2020
\(\left(\frac{x-10}{1994}-1\right)\)+\(\left(\frac{x-8}{1996}-1\right)\)+\(\left(\frac{x-6}{1998}-1\right)\)+\(\left(\frac{x-4}{2000}-1\right)\)+\(\left(\frac{x-2}{2002}-1\right)\)=\(\left(\frac{x-2002}{2}-1\right)\)+\(\left(\frac{x-2000}{4}-1\right)\)+\(\left(\frac{x-1998}{6}-1\right)\)+\(\left(\frac{x-1996}{8}-1\right)\)+\(\left(\frac{x-1994}{10}-1\right)\)
suy ra \(\frac{x-2004}{1994}\)+\(\frac{x-2004}{1996}\)+\(\frac{x-2004}{1998}\)+\(\frac{x-2004}{2000}\)+\(\frac{x-2004}{2002}\)=\(\frac{x-2004}{2}\)+\(\frac{x-2004}{4}\)+\(\frac{x-2004}{6}\)+\(\frac{x-2004}{8}\)+\(\frac{x-2004}{10}\)
suy ra \(\frac{x-2004}{1994}\)+\(\frac{x-2004}{1996}\)+\(\frac{x-2004}{1998}\)+\(\frac{x-2004}{2000}\)+\(\frac{x-2004}{2002}\)- \(\frac{x-2004}{2}\)- \(\frac{x-2004}{4}\)- \(\frac{x-2004}{6}\)- \(\frac{x-2004}{8}\)- \(\frac{x-2004}{10}\)=0
suy ra (x-2004) . ( \(\frac{1}{1994}\)+\(\frac{1}{1996}\)+\(\frac{1}{1998}\)+\(\frac{1}{2000}\)+\(\frac{1}{2002}\)-\(\frac{1}{2}\)-\(\frac{1}{4}\)-\(\frac{1}{6}\)- \(\frac{1}{8}\)- \(\frac{1}{10}\))=0
Vì \(\frac{1}{1994}\)+\(\frac{1}{1996}\)+\(\frac{1}{1998}\)+\(\frac{1}{2000}\)+\(\frac{1}{2002}\)-\(\frac{1}{2}\)-\(\frac{1}{4}\)-\(\frac{1}{6}\)- \(\frac{1}{8}\)- \(\frac{1}{10}\) khác 0
nên x-2004=0 suy ra x=2004
:)))))))))))))))))))))))))))))))))))))))))))
23 tháng 1
2+(-4)+............+(-x)=-2000
2+(-4)]+...........+[(x-2)+(-x)]=-2000
(-2)+..............+(-2)=-2000(có x:2 số -2)
(-2).(x:2)=-2000
x:2=(-2000):(-2)
x:2=1000
x=1000.2
x=2000
Đặt A= [2+(-4)]+[6+(-8)]+...+(x-2+x)=2000
Ta có:A= (-2)x {[(x-2):2+1]:2}=2000
A=(-2)x[(x-2):2+1]=2000x2
(x-2):2+1 =4000:(-2)
(x-2) =(-2000-1)x2
x =-4002+2=-4000