Nghiem Cua Da Thuc Mot Bien 7

  • June 2020
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KiÓm tra bµi cò TÝnh : f(x) + g(x) – h(x) biÕt : f(x) = x5 – 4x3 + x2 – 2x + 1 g(x) = x5 – 2x4 + x2 – 5x + 3 h(x) = x4 – 3x2 + 2x – 5 Gäi ®a thøc f(x) + g(x) – h(x) lµ A(x). TÝnh A(1) ? Gi¶i f(x) = x5

+

– 4x3 + x2 – 2x + 1

g(x) = x5 – 2x4 - h(x) =

- x4

+ x2 – 5x + 3 + 3x2 - 2x + 5

f(x) + g(x) - h(x) = 2x5 – 3x4 – 4x3 + 5x2 - 9x +9 A(x) = f(x) + g(x) - h(x) = 2x5 – 3x4 – 4x3 + 5x2 - 9x + 9 A(1) = 2 . 15 – 3 . 14 – 4 . 13 + 5 . 12 – 9 . 1 + 9 =2.1–3.1–4.1+5.1–9+9

TiÕt 62

: §9.NghiÖm cña ®a thøc mét biÕn

1) NghiÖm cña ®a thøc mét biÕn a. XÐt bµi to¸n : Cho biÕt c«ng thøc ®æi tõ ®é F sang ®é C lµ : 5 C=

9

( F − 32)

Hái n­íc ®ãng b¨ng ë bao nhiªu ®é F ? b. Kh¸i niÖm : NÕu t¹i x = a, ®a thøc P(x) cã gi¸ trÞ b»ng 0 th× ta nãi a ( hoÆc x = a ) lµ mét nghiÖm cña ®a thøc ®ã.

TiÕt 62

: §9.NghiÖm cña ®a thøc mét biÕn

2) VÝ dô a. Cho ®a thøc P(x) = 2x + 1

1 TÝnhP (− ) 2

?

1  1 P(− ) = 2. −  + 1 = −1 + 1 = 0 2  2

1

⇒ x=− +1 2

lµ nghiÖm cña ®a thøc P(x) = 2x

TiÕt 62

: §9.NghiÖm cña ®a thøc mét biÕn

2) VÝ dô a. Cho ®a thøc P(x) = 2x + 1

1 x=− + 12

lµ nghiÖm cña ®a thøc P(x) = 2x

b. Cho ®a thøc Q(x) = x2 - 1

T¹i sao x = 1, x = -1 lµ nghiÖm cña ®a thøc Q(x) ? Q(x) cã nghiÖm lµ 1 vµ - 1 v× : Q (1) = 12 – 1 = 1 – 1 = 0 Q(-1) = (- 1)2 – 1 = 1 – 1 = 0

TiÕt 62

: §9.NghiÖm cña ®a thøc mét biÕn

2) VÝ dô a. Cho ®a thøc P(x) = 2x + 1

1 x=− + 12

lµ nghiÖm cña ®a thøc P(x) = 2x

b. Cho ®a thøc Q(x) = x2 - 1 x = 1, x = -1 lµ nghiÖm cña ®a thøc Q(x) = x2 - 1 c. Cho ®a thøc G(x) = x2 + 1

T×m nghiÖm cña ®a thøc G(x) ? G(x) kh«ng cã nghiÖm v× : mäi x

x2 ≥ 0 víi mäi x ⇒ x2 + 1 > 0 víi

TiÕt 62

: §9.NghiÖm cña ®a thøc mét biÕn

2) VÝ dô a. Cho ®a thøc P(x) = 2x + 1

1 x=− + 12

lµ nghiÖm cña ®a thøc P(x) = 2x

b. Cho ®a thøc Q(x) = x2 - 1 x = 1, x = -1 lµ nghiÖm cña ®a thøc Q(x) = x2 - 1 c. Cho ®a thøc G(x) = x2 + 1 §a thøc G(x) kh«ng cã nghiÖm Chó ý : - Mét ®a thøc ( kh¸c ®a thøc kh«ng ) cã thÓ cã mét nghiÖm, hai nghiÖm, ... hoÆc kh«ng cã nghiÖm. - Sè nghiÖm cña mét ®a thøc ( kh¸c ®a thøc kh«ng ) kh«ng v­ît qu¸ bËc cña nã.

? x = - 2, x = 0 vµ x = 2 cã ph¶i lµ c¸c 1 nghiÖm cña ®a thøc x3 – 4x hay kh«ng ? V× sao ? Gi¶i Gäi P(x) = x3 – 4x P(-2) = (-2)3 – 4.(-2) = - 8 – (- 8) = -8 + 8 = 0 P(0) = 03 – 4.0 = 0 – 0 = 0 P(2) = 23 – 4.2 = 8 – 8 = 0 VËy x = - 2, x = 0, x = 2 lµ nghiÖm cña ®a thøc P(x) = x3 – 4x

Muèn kiÓm tra mét sè a cã lµ nghiÖm cña ®a thøc P(x) hay kh«ng ta lµm nh­ sau : - TÝnh P(a). - KÕt luËn : + NÕu P(a) = 0 Th× x = a lµ mét nghiÖm cña ®a thøc P(x) + NÕu P(a) ≠ 0 Th× x = a kh«ng lµ nghiÖm cña ®a thøc P(x)

? Trong c¸c sè cho sau mçi ®a thøc, sè nµo 2 lµ nghiÖm cña ®a thøc ? 1 1 1 1 a) P( x) = 2 x +

2

b) Q(x) = x2 – 2x 3

4

2

3

1



4

-1

Gi¶i a) Cho P(x) = 0

2x +

1 =0 2

1 2 1 x =- :2 2 1 x = 4 VËy nghiÖm cña ®a thøc 2x

= -

b) Q(x) = x2 – 2x – 3 Q(3) = 32 - 2.3 – 3 = 9 – 6 – 3 = 3 – 3 =0 Q(1) = 12 – 2.1 – 3 = 1 – 2 – 3 = - 1 – 3 =-4≠ 0 Q(-1) = (-1)2 – 2.(-1) – 3 = 1 + 2 – 3 = 3–3=0 VËy x = 3, x = - 1 lµ nghiÖm cña ®a thøc

1 4

P(x)− lµ x =

Q(x) = x2

Trß ch¬i

Cho ®a thøc P(x) = x3 – x Trong c¸c sè sau : - 2, - 1, 0, 1, 2 sè nµo lµ nghiÖm cña ®a thøc P(x)

H­íng dÉn vÒ nhµ - Häc lý thuyÕt. - Lµm bµi tËp : 54, 56 ( SGK Tr 48 ) 43, 44, 46 ( SBT Tr 15 )

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