Question

Phân tích đa thức thành nhân tử :

a, a\(^2\) - 4b\(^2\)

b, 81( x + 7 )\(^2\) - ( 3x + 8 )\(^2\)

c, 4 - 4x + x\(^2\) - 4y\(^2\) - 4y - 1

d, y\(^2\) - 4xy + 4x\(^2\) - 16

e, 9x\(^2\) - 6x + 1 - (3x)\(^2\)

f, x\(^3\) + 3x\(^2\)y + 3xy\(^2\) + y\(^3\) + x + y

Tìm x, biết :

a, x\(^3\) + 27 + ( x + 3 ).( x - 9 ) = 0

b, x\(^4\) - 6x\(^3\) + 12x\(^2\) - 8x = 0

c, 36( x - 1 )\(^2\) - 25( 2x - 1 )\(^2\) = 0

Answer 1

a) a² − 4b² 

= a² − (2b)²
= (a −2b)(a + 2b)

Answer 2

b) 81(x+ 7)² - (3x + 8)²

= [9(x + 7)]² - (3x + 8)²

= (9x + 63)² - (3x + 8)²

= (9x + 63 - 3x - 8)(9x + 63 +3x + 8)
= (6x + 55)(12x + 71)

Answer 3

c) 4 - 4x + x² - 4y² - 4y - 1

= (x² - 4x + 4) - (4y² + 4y + 1)

= (x - 2)² - (2y + 1)²

= (x - 2 - 2y - 1)(x - 2 + 2y + 1)

= (x - 2y - 3)(x + 2y - 1)

Answer 4

d) y² - 4xy + 4x² - 16

= (y² - 4xy + 4x²) - 16

= (y - 2x)² - 4²

= (y - 2x - 4)(y - 2x + 4)

Answer 5

e) 9x² - 6x + 1 - (3x)²

= (9x² - 6x + 1) - (3x)²

= (3x - 1)² - (3x)²

= (3x - 1 - 3x)(3x - 1 + 3x)
= -(6x - 1)
= 1 - 6x

Answer 6

f) x³ + 3x²y + 3xy² + y³ + x + y

= (x³ + 3x²y + 3xy² + y³) + (x + y)

= (x + y)³ + (x + y)

= (x + y)[(x + y)² + 1]
= (x + y)(x² + 2xy + y² + 1)

Answer 7

Tìm x

a) x³ + 27 + (x + 3)(x - 9) = 0

(x³ + 27) + (x + 3)(x - 9) = 0

(x + 3)(x² - 3x + 9) + (x + 3)(x - 9) = 0

(x + 3)(x² - 3x + 9 + x - 9) = 0

(x + 3)(x² - x) = 0

x(x + 3)(x - 2) = 0
x = 0 hoặc x + 3 = 0 hoặc x - 2 = 0
*) x + 3 = 0
x = -3

*) x - 2 = 0
x = 2
Vậy x = -3; x = 0; x = 2

Answer 8

Tìm x
b) x⁴ - 6x³ + 12x² - 8x = 0

x(x³ - 6x² + 12x - 8) = 0

x(x - 2)³ = 0

x = 0 hoặc (x - 2)³ = 0

*) (x - 2)³ = 0

x - 2 = 0
x = 2
Vậy x = 0; x = 2

Answer 9

Tìm x
c) 36(x - 1)² - 25(2x - 1)² = 0

[6(x - 1)]² - [5(2x - 1)]² = 0

(6x - 6)² - (10x - 5)² = 0

(6x - 6 - 10x + 5)(6x - 6 + 10x - 5) = 0
(-4x - 1)(16x - 11) = 0
-4x - 1 = 0 hoặc 16x - 11 = 0

Vậy: