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a)
200 - ( 2 . x + 6 ) = 8 . 2
200 - ( 2 . x + 6 ) = 16
2 . x + 6 = 200 - 16
2 . x + 6 = 184
x + 6 = 184 : 2
x + 6 = 92
x = 92 - 6
x = 86
b)
2 . x : 4 = 16
x : 4 = 16 : 2
x : 4 = 8
x = 8 . 4
x = 32
c)
2008 . x : 2008 . 5 = 2008 . 3
2008. ( x : 5 ) = 6024
x : 5 = 6024 : 2008
x : 5 = 3
x = 3 . 5
x = 15
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Vì (2x - 1)2008 \(\ge\) 0 với mọi x
(y - \(\frac{2}{5}\))2008 \(\ge\) 0 với mọi y
|x + y - z| \(\ge\) 0 với mọi x; y ;z
=> (2x-1)2008+(y-\(\frac{2}{5}\))2008+|x+y-z| \(\ge\) 0 với mọi x; y ;z
Để (2x-1)2008+(y-\(\frac{2}{5}\))2008+|x+y-z| = 0
<=> (2x-1)2008 = 0 ; (y-\(\frac{2}{5}\))2008 = 0 ; |x+y-z| = 0
=> 2x -1 = 0 ; y - \(\frac{2}{5}\)= 0 ; x+ y - z = 0
=> x = \(\frac{1}{2}\) ; y = \(\frac{2}{5}\) ; z = x + y = \(\frac{1}{2}\) + \(\frac{2}{5}\) = \(\frac{9}{10}\)
KL:...
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(x+2007) + ( x+1+2006) + ..... +0 =0
=> x +2007 =0
=> x =-2007
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x+2007) + ( x+1+2006) + ..... +0 =0
=> x +2007 =0
=> x =-2007
![](https://rs.olm.vn/images/avt/0.png?1311)
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Ta có : \(5+3x+1=86\)
\(\Leftrightarrow3x=86-5-1=80\)
\(\Leftrightarrow x=\frac{80}{3}\)
Vậy \(x=\frac{80}{3}\)
\(2008^x:2008^5=2008^3\)
\(\Leftrightarrow2008^{x-5}=2008^3\)
\(\Leftrightarrow x-5=3\Leftrightarrow x=8\)
Vậy \(x=8\)
\(6^x:12=3\)
\(\Leftrightarrow6^x=12\cdot3=36\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)