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=> x + (x+1)+.....+ 2003+2004 =0
=>(x+2004) +( x+1+2003) + .....=0 ( giả sử có k dấu ngoặc)
=> k ( x+2004) = 0
=> x+ 2004 =0
=> x =-2004
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
A = 2005 x 2005 + 1 / 2005 x 2005 x 2005 - 1
A = 2005 x 2005 + 1 / 2005 x 2005 x 2005 - 1
A = \(\frac{2005+1}{2005x2005-1}\)
B = \(\frac{2005+1}{2005x2005-1}\)
=> A = B
a)x-2006=1=>x=2006
b)(x+50)*50-50=2500
(x+50)*50=2550
x+50=51
x=1
Tick Nha
2005.(x-2006) = 2005
x - 2006 = 1
x = 2007
[(x+50).50-50]:50 = 50
(x+50).50-50 = 50 x 50 = 2500
(x+50).50 = 2500 + 50 = 2550
x + 50 = 2550 : 50 = 51
x = 1
(x - 2005) x 2006 = 0
<=> x - 2005 = 0
<=> x = 2005
Vậy x = 2005
\(456+\left(x-357\right)=1362\) \(\left(2345-x\right)-183=2014\)
\(x-357=906\) \(2345-x=2197\)
\(x=1263\) \(x=148\)
\(\left(x-2005\right).2006=0\) Phần sau tương tự thế !
\(x-2005=0\)
\(x=2005\)
`#3107.101107`
A,
\(2\times32\times12+4\times6\times41+8\times27\times3\\ =24\times32+24\times41+24\times27\\ =24\times\left(32+41+27\right)\\ =24\times100\\ =2400\)
B,
\(\left(2006\times2005^{2016}-2005^{2016}\right)\div2005^{2017}\\ =\left[2005^{2016}\times\left(2006-1\right)\right]\div2005^{2017}\\ =\left(2005^{2016}\times2005\right)\div2005^{2017}\\ =2005^{2017}\div2005^{2017}\\ =1\)
=> x . (2005 - 1) = 2005 . (2003 + 1)
=> x . 2004 = 2005 . 2004
=> x = 2005