5 gdz algebra makar

Page 160

340.

o2 = t

Z

H[hagZqbf 2 D = 4 − 4 ⋅ 4 ⋅ (− 15) = 256 ;

t1 =

⇒ x 2 = 1,5 beb x 2 = −2,5

[ Imklv o

−4 + 16 = 1,5 8

g_l dhjg_c

beb

4t 2 + 4t − 15 = 0 ; −4 − 16 t2 = = −2,5 8

x1 = 1,5

beb

x 2 = − 1,5

= t ⇒ 2t − t − 36 = 0 ; D = (−1) − 4 ⋅ 2 ⋅ (− 36) = 289 ; 1 + 17 1 − 17 t1 = = 4,5 beb t 2 = = −4 ⇒ x 2 = 4,5 beb x 2 = −4 g_l 4 4

dhjg_c

2

2

x1 = 4,5

beb

2

x 2 = − 4,5

341.

(

Z

)(

)

1 3 −6 1 a b ⋅ 3a − 2b 5 = ⋅ 3 a 3 ⋅ a − 2 b − 6 ⋅ b 5 = 2 2 3 3 − 2 − 6 + 5 3 −1 3a = a ⋅b = ab = 2 2 2b 1 − 3 3 [ 3a − b ⋅ (4ab ) = 3a − b ⋅ 4 −1 ⋅ a −1 ⋅ b −1 = 3 a −3 a −1 bb −1 = 3 a − 4 . 4 4 \ a–6b10(2a–2b4)–2= 4a −6b10 ⋅ 2−2 ⋅ a 4 ⋅ b−8

(

=

(

)(

]

10ab −5

)( )

)

4 −6 4 10 −8 a a b b = a −6+4 ⋅ b10−8 = a −2b 2 4 1 33

−2 3

a b

=

( )(

)

10 ⋅ 3 aa 2 b − 5b − 3 = 3a1+ 2 ⋅ b − 5 + (− 3) = 3a 3b −8 . 10

342.

Z 81 ⋅ 3− 6 = 34 ⋅ 3− 6 = 32 = 34 − 6 = 3− 2 = 1 [

(− 3 )

−3 3

−9

−2

\ 9 −5 ⋅  1  9

]

9

=

(− 3) 4 − (3)−

−3

(− 3 ) ⋅ 27 −3 2

−9

=

3

4

3

9

= 3 4 −9 = 3 −5 =

( ) ⋅ (3 )

= 32 3

−5

− 2 −3

= 3 −10 ⋅ 3 6 = 3 − 4 =

( ) =3

= (− 3) ⋅ 3 −6

3 3

1 . 243

−6

⋅3 = 3 = 3 9

3

1 3

4

−6+9

=

1 . 81

= 33 = 27.

53

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