ГДЗ.+Алгебра+и+начала+анализа.+10-11кл_10+класс_Алимов%2C+Ко

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43 + 432 − 1800 43 − 432 − 1800 + + 2 2

43 + 30 2 + 43 − 30 2 =

109.

43 + 432 − 1800 43 − 432 − 1800 43 − 1849 − 1800 43 + 49 − =2 =2 = 2 2 2 2

+ =2

43 + 7 50 =2 = 2 25 = 10 . 2 2

110. a = (4 − 3 2)2 + 8 34 − 24 2 − 5 = 16 − 24 2 + 18 +  34 + 1156 − 1152  34 + 1156 − 1152 + 8 − 2 2  

   − 5 = 34 − 

− 24 2 + 8 ⋅ (3 2 − 4) − 5 = 34 − 24 2 + 24 2 − 32 − 5 = = 2 − 5 ; 2 − 5 < 0 , так как 2 < 5 , значит, a < 0 . 2

111. 1) a = b=

5− 3

2

+

5 3+ 2 2

2

;

5− 3

2) a = 2 + 3 ; значит, a < b ;

2 < 1,4143;

4) a = 13 − 12 ; 13 < 3,604; a < 0,14 < 0,147 < b .

=

2 2− 3

5

=

5 + 10

=

2( 2 + 3 )

5 (5 − 10 ) (5 − 10 )(5 + 10 )

=

17 < 4,124; b < 1,124 < 1,127 < a ,

12 > 3,464;

( 2 − 3 )( 2 + 3 )

5 5 ( 5 )2 ⋅ 2 5 5 − 5 2 = = 15 15

=

4⋅ 2

5) =

3 4

5 −4 2

=

3

8

3

11 < 3,317;

2( 2 + 3 ) = −2( 2 + 3 ) ; 2−3

5 (5 − 10 ) = 25 − 10

5 (5 − 10 ) = 15

5− 2 ; 3

3 3 3 3 3) 3 = 3 ⋅ 2 = 3 ⋅ 2 = 3 ⋅ 2 = 3 ⋅ 2 ; 4) 3 3 3

4

> 0,8;

3 < 1,7321; a < 3,1464 < 3,1622 < 10 = b,

3) a = 5 − 5 ; 15 < 3,873; a > 1,127; значит, b < a ;

2)

3+ 2 2

< 3,4; значит, b < 3,4 < 4,7 < a , значит, b < a;

8− 5

112. 1)

5

> 3,9;

2

(2) 3

3( 4 5 + 4 2 ) ( 4 5 − 4 2 )( 4 5 + 4 2 )

=

3( 4 5 + 4 2 ) 5− 2

2 4

27

=

24 3 4

27 ⋅ 4 27

=

=

3( 4 5 + 4 2 )( 5 + 2 ) 3( 4 5 + 4 2 )( 5 + 2 ) = ( 4 5 + 4 2 )( 5 + 2 ) ; = 5−2 3

34

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24 3 4

34

=

24 3 ; 3


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