Question

1.Phân tích các đa thức sau thành nhân tử:

a, \(x^2+6x+9\)

b, \(10x-25-x^2\)

c, \(8x^3-\dfrac{1}{8}\)

d, \(\dfrac{1}{25}x^2-64y^2\)

2.Phân tích các đa thức sau thành nhân tử:

a, \(x^3_{ }+\dfrac{1}{27}\)

b, \(\left(a+b\right)^3-\left(a-b\right)^3\)

c, \(\left(a+b\right)^3+\left(a-b\right)^3\)

d, \(8x^3+12x^2y+6xy^2+y^3\)

e, \(-x^3+9x^2-27x+27\)

3.Tìm \(x\),biết:

a, \(2-25x^2=0\)

b, \(x^2-x+\dfrac{1}{4}=0\)

Answer 1

Bài giải:

1.

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

= -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - = (2x)³ – ()³ = (2x - )[(2x)² + 2x . + ()²]

^{= (2x - )(4x2 + x + )}

d) x² – 64y² = - (8y)² = (x + 8y)(x - 8y)

2.

a) x³ + $\frac{1}{27}$ = x³ + ($\frac{1}{3}$)³ = (x + $\frac{1}{3}$)(x² – x . $\frac{1}{3}$+ ($\frac{1}{3}$)²)

=(x + $\frac{1}{3}$)(x² – $\frac{1}{3}$x + $\frac{1}{9}$)

b) (a + b)³ – (a - b)³

= [(a + b) – (a – b)][(a + b)² + (a + b) . (a – b) + (a – b)²]

= (a + b – a + b)(a² + 2ab + b² + a² – b² + a² – 2ab + b²)

= 2b . (3a³ + b²)

c) (a + b)³ + (a – b)³ = [(a + b) + (a – b)][(a + b)² – (a + b)(a – b) + (a – b)²]

= (a + b + a – b)(a² + 2ab + b² – a² +b² + a² – 2ab + b²]

= 2a . (a² + 3b²)

d) 8x³ + 12x²y + 6xy² + y³ = (2x)³ + 3 . (2x)² . y +3 . 2x . y + y³ = (2x + y)³

e) - x³ + 9x² – 27x + 27 = 27 – 27x + 9x² – x³ = 3³ – 3 . 3² . x + 3 . 3 . x² – x³ = (3 – x)³