1 2 A      A 1 A 1 A (A 1) (A 1)          2 2 2( A 1) 1 2 A ( A 1) ( A 1)=

Question

Câu 2: a) A =  a 1 2 a : 1 2 a      a 1 a 1 a (a 1) (a 1)          2 2 2( a 1) 1 2 a ( a 1) ( a 1)= : :         . a 1 a 1 (a 1)( a 1) a 1 ( a 1)(a 1)  2( a 1) (a 1)( a 1) . a 1  . a 1 ( a 1)b) a = 2011 - 2 2010  ( 2010  1 )2 a  2010  1Vậy A = 2010 . 1 ta có:

Answer

Câu 2: a) A =  a 1 2 a : 1 2 a      a 1 a 1 a (a 1) (a 1)          2 2 2( a 1) 1 2 a ( a 1) ( a 1)= : :         . a 1 a 1 (a 1)( a 1) a 1 ( a 1)(a 1)  2( a 1) (a 1)( a 1) . a 1  . a 1 ( a 1)b) a = 2011 - 2 2010  ( 2010  1 )2 a  2010  1Vậy A = 2010 . 1 ta có:

1 2 A      A 1 A 1 A (A 1) (A 1)          2 2 2( A 1) 1 2 A ( A 1) ( A 1)=

Câu 2: a) A = ^{a 1 2 a} : ¹ ^{2 a}

      

a 1 a 1 a (a 1) (a 1)

   

 

      

( a 1) 1 2 a ( a 1) ( a 1)

=

: :

         .

a 1 a 1 (a 1)( a 1) a 1 ( a 1)(a 1)

  

( a 1) (a 1)( a 1)

 

. a 1

  .

a 1 ( a 1)

b) a = 2011 - 2 2010  ( 2010  1 )

 a  2010  1

Vậy A = 2010 .

1 ta có:

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