=> 2x - x = 2 - 3

=> x = -1

With 2x + 3 = -(x + 2)

=> 2x + 3 = -x - 2

=> 2x + x = -2 - 3

=> 3x = -5

=> x = $\frac{-5}{3}$

So x = -1 and x = $\frac{-5}{3}$

b) We have : 

A = |x - 2006| + |2007 - x| $\ge$ |x - 2006 + 2007 - x| = |1| = 1

$\iff \{\begin{array}{c}\left|x-2006\right|\le 0\\ \left|2007-x\right|\le 0\end{array}\Rightarrow \{\begin{array}{c}x=2006\\ x=2007\end{array}$

So when x = 2006 ; 2007 then value of A smallest 

=> 2x - x = 2 - 3

=> x = -1

With 2x + 3 = -(x + 2)

=> 2x + 3 = -x - 2

=> 2x + x = -2 - 3

=> 3x = -5

=> x = $\frac{-5}{3}$

So x = -1 and x = $\frac{-5}{3}$

b) We have : 

A = |x - 2006| + |2007 - x| $\ge$ |x - 2006 + 2007 - x| = |1| = 1

$\iff \{\begin{array}{c}\left|x-2006\right|\le 0\\ \left|2007-x\right|\le 0\end{array}\Rightarrow \{\begin{array}{c}x=2006\\ x=2007\end{array}$

So when x = 2006 ; 2007 then value of A smallest 

=> 2x - x = 2 - 3

=> x = -1

With 2x + 3 = -(x + 2)

=> 2x + 3 = -x - 2

=> 2x + x = -2 - 3

=> 3x = -5

=> x = $\frac{-5}{3}$

So x = -1 and x = $\frac{-5}{3}$

b) We have : 

A = |x - 2006| + |2007 - x| $\ge$ |x - 2006 + 2007 - x| = |1| = 1

$\iff \{\begin{array}{c}\left|x-2006\right|\le 0\\ \left|2007-x\right|\le 0\end{array}\Rightarrow \{\begin{array}{c}x=2006\\ x=2007\end{array}$

So when x = 2006 ; 2007 then value of A smallest 

We have : ( x + 1 ) . ( 1 + 1999 ) = 4000

 (x + 1 ) . 2000 = 4000

   x + 1 = 4000 : 2000

   x + 1 = 2 

   x = 2 - 1 

  x = 1 

We have : ( x + 1 ) . ( 1 + 1999 ) = 4000

 (x + 1 ) . 2000 = 4000

   x + 1 = 4000 : 2000

   x + 1 = 2 

   x = 2 - 1 

  x = 1 

We have : ( x + 1 ) . ( 1 + 1999 ) = 4000

 (x + 1 ) . 2000 = 4000

   x + 1 = 4000 : 2000

   x + 1 = 2 

   x = 2 - 1 

  x = 1 

We have : (x + 2)(x + 2) = 24(x + 2)(x + 2) = 24

⇒x2 + 4x + 4 =24

=> x2 + 4x = 20

=> x(x + 4) = 20

=> x ,x + 4 thuộc Ư(20) = {1;2;4;5;10;20}

 When x = 1 thì x + 4 = 20 => x = 16 

Not yes value satisfies

We have : (x + 2)(x + 2) = 24(x + 2)(x + 2) = 24

⇒x2 + 4x + 4 =24

=> x2 + 4x = 20

=> x(x + 4) = 20

=> x ,x + 4 thuộc Ư(20) = {1;2;4;5;10;20}

 When x = 1 thì x + 4 = 20 => x = 16 

Not yes value satisfies

We have : (x + 2)(x + 2) = 24(x + 2)(x + 2) = 24

⇒x2 + 4x + 4 =24

=> x2 + 4x = 20

=> x(x + 4) = 20

=> x ,x + 4 thuộc Ư(20) = {1;2;4;5;10;20}

 When x = 1 thì x + 4 = 20 => x = 16 

Not yes value satisfies

Number of female students is :

  40 x 2525= 16 ( student )

Number of male students is :

   40 - 16 = 24 ( student )

      Answer : 24 student

Number of female students is :

  40 x 2525= 16 ( student )

Number of male students is :

   40 - 16 = 24 ( student )

      Answer : 24 student

Number of female students is :

  40 x 2525= 16 ( student )

Number of male students is :

   40 - 16 = 24 ( student )

      Answer : 24 student

To be able to buy 9 bottles of fruit juice at the cheapest price, they should appoint 4 people to buy 8 bottles for: (7 + 1) x 4 = 32 (dollars)

The total amount they will pay to buy 9 bottles is: 32 + 7 = 39 (dollars)

To be able to buy 9 bottles of fruit juice at the cheapest price, they should appoint 4 people to buy 8 bottles for: (7 + 1) x 4 = 32 (dollars)

The total amount they will pay to buy 9 bottles is: 32 + 7 = 39 (dollars)

To be able to buy 9 bottles of fruit juice at the cheapest price, they should appoint 4 people to buy 8 bottles for: (7 + 1) x 4 = 32 (dollars)

The total amount they will pay to buy 9 bottles is: 32 + 7 = 39 (dollars)

Number of female students is:         =20 ( students )

Number of male students is :   50 - 20 = 30 ( students )

                                               Answer : 30 students

Number of female students is:         =20 ( students )

Number of male students is :   50 - 20 = 30 ( students )

                                               Answer : 30 students

Number of female students is:         =20 ( students )

Number of male students is :   50 - 20 = 30 ( students )

                                               Answer : 30 students

Number of male students is :   50 - 20 = 30 ( students )

                                               Answer : 30 students

We have : 

2a+b+c+da=a+2b+c+db=a+b+2c+dc=a+b+c+2dd2a+b+c+da=a+2b+c+db=a+b+2c+dc=a+b+c+2dd

⇒2a+b+c+da−1=a+2b+c+db−1=a+b+2c+dc−1=a+b+c+2dd−1⇒2a+b+c+da−1=a+2b+c+db−1=a+b+2c+dc−1=a+b+c+2dd−1

⇒a+b+c+da=a+b+c+db=a+b+c+dc=a+b+c+dd⇒a+b+c+da=a+b+c+db=a+b+c+dc=a+b+c+dd

+> If a + b + c + d ≠0≠0

=> a = b = c = d (Because the same of namerator)

=> M = 1 + 1 + 1 + 1 = 4

+> If a + b + c + d = 0

=> a + b = -(c + d)

      b + c = -(a + d)

      c + d = -(a + b)

      a + d = -(b + c)

Change this into M , we have :

M=a+bc+d+b+cd+a+c+da+b+a+db+cM=a+bc+d+b+cd+a+c+da+b+a+db+c

M=−(c+d)c+d+−(a+d)a+d+−(a+b)a+b+−(b+c)b+c=(−1)+(−1)+(−1)+(−1)=−4M=−(c+d)c+d+−(a+d)a+d+−(a+b)a+b+−(b+c)b+c=(−1)+(−1)+(−1)+(−1)=−4

So M = 4 and M = -4