|x + 2006| = 0
Cho . Tính giá tri của biểu thức:
x = 2005
=> x + 1 = 2006
Đặt A = x2005 - 2006x2004 + 2006x2003 - 2006x2002 + .... - 2006x2 + 2006x - 1
= x2005 - (x + 1)x2004 + (x + 1)x2003 - (x + 1)x2002 + .... - (x + 1)x2 + (x + 1)x - 1
= x2005 - x2005 - x2004 + x2004 + x2003 - x2003 - x2002 + ... - x3 - x2 + x2 + x - 1
= x - 1
= 2005 - 1 = 2004
Vậy A = 2004
tim x
a) 3x(x-2006)-x+2006=0
b) x=9x^3
a)3x(x-2006)-x+2006=0
<=>3x(x-2006)-(x-2006)=0
<=>(x-2006)(3x-1)=0
<=>x-2006=0 or 3x-1=0
<=>x=2006 or x=1/3
b)x=9x3
<=>9x3-x=0
<=>x(9x2-1)=0
<=>x(3x-1)(3x+1)=0
<=>x=0 or 3x-1=0 or 3x+1=0
<=>x=0 or x=1/3 or x=-1/3
Giải PT sau: \(x^2-2006\left[x\right]-x+2006=0\)
[x] la tri tuyet doi ha phuc
khong biet phuc do hay la hoi nua.
TH1: x>=0
x^2-2006x-x+2006=0
x(x-1)-2006(x-1)=0
(x-1)(x-2006)=0
x=1 (nhan)
x=2006 (nhan)
TH2: x<0
x^2+2006x-x+2006=0
x^2+2005x+2006=0
delta=2005^2-4.2006=2005^2-4.(2005+1)=2005^2-4.2005-4=(2005-2)^2-8=2003^2-8 le qua ke no thoi (hay sai o dau do)
can (delta)~~2002<2005 nhan het
x=(-2005+-can(delta)/2
Bài 1 : cho 3 số a , ,y, x lớn hơn hoặc bằng 0 và x^2006 + y ^2006 + z^2006 = 3 . Tìm GTLN của A = x ^ 2 + y^2 + z^2 .
(195-13.15):(1945+1014)
(x-2005).2006=0
480+45.4=(x+125):5+260
2005.(x-2006)=2005
[(x+50).50-50]:50=50
nhầm
câu c :
=> 480 + 180 = (x + 125) : 5 + 260
=> 660 = (x + 125) : 5 + 260
=> (x + 125) : 5 = 400
=> x + 125 = 2000
=> x = 1875
câu e :
=> (x + 50) . 50 - 50 = 2500
=> (x + 50) . 50 = 2550
=> x + 50 = 51
=> x = 1
a/ = (195 - 195) : (1945 + 1014) = 0 : (1945 + 1014) = 0
b/ => x - 2005 = 0 => x = 0 + 2005 = 2005
c/ => 480 + 180 = (x + 125) : 5 + 260
=> 300 = (x + 125) : 5 + 260
=> (x + 125) : 5 = 40
=> x + 125 = 8
=> x = -117
d/ => x - 2006 = 1 => x = 2007
e/ => (x + 50) . 50 - 50 = 1
=> (x + 50) . 50 = 51
=> x + 50 = 51/50
=> x = -2449/50
Tìm x
(x-2005).2006=0
480+45.4=(x+125):5+260
2005.(x-2006)=2005
{(x+50).50-50}:50=50
a)(x-2005).2006=0
=>x-2005=0
=>x=2005
b)480+45.4=(x+125):5+260
=>480+180=(x+125):5+260
=>660=(x+125):5+260
=>(x+125):5=400
=>x+125=2000
=>x=1875
c)2005.(x-2006)=2005
=>x-2006=1
=>x=2007
d){(x+50).50-50}:50=50
=>(x+50).50-50=2500
=>(x+50)*50=2550
=>x+50=51
=>x=1
(x-2005).2006=0. Tìm x
(x-2005).2006=0
(x-2005)=0:2006
(x-2005)=0
x=0+2005
=2005
a (x-2005). 2006=0
b 2005.(x-2006)=2005
c 480+45.4=(x+125):5+260
d [(x+50).50-50]:50=50
a) (x - 2005 ) . 2006 = 0
x - 2005 = 0 : 2006
x - 2005 = 0
x = 0 + 2005
x = 2005
b) 2005.(x-2006) = 2005
x - 2006 = 2005 : 2005
x - 2006 = 1
x = 1 + 2006
x = 2007
c) 480+45.4=(x+125) : 5 + 260
480 + 180 = (x+125) : 5 + 260
660 = ( x + 125 ) : 5 + 260
660 - 260 = ( x+ 125 ) : 5
400 = ( x + 125 ) : 5
400 x 5 = x + 125
2000 = x + 125
2000 - 125 = x
1875 = x
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