Question

Phân tích đa thức thành nhân tử.

Answer 1

b)x³-2x²-4xy²+x

=x(x²-2x-4y²+1)

=x[(x²-2x+1)-4y²]

=x[(x-1)²-4y²]

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

Answer 2

c) (x+2)(x+3)(x+4)(x+5)-8

=[(x+2)(x+5)][(x+3)(x+4)]-8

=(x²+5x+2x+10)(x²+4x+3x+12)-8

=(x²+7x+10)(x²+7x+12)-8

đặt x²+7x+10 =a ta có

a(a+2)-8

=a²+2a-8

=a²+4a-2a-8

=(a²+4a)-(2a+8)

=a(a+4)-2(a+4)

=(a+4)(a-2)

thay a=x²+7x+10 ta đc

(x²+7x+10+4)(x²+7x+10-2)

=(x²+7x+14)(x²+7x+8)

bài 2 x³-x²y+3x-3y

=(x³-x²y)+(3x-3y)

=x²(x-y)+3(x-y)

=(x-y)(x²+3)

Answer 3

a) x³ - 2x² - 4xy² + x

= x(x² - 2x - 4y² + 1)

= x[(x² - 2x + 1) - 4y²]

= x[(x - 1)² - 4y²]

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

b) (x + 2)(x + 3)(x + 4)(x + 5) - 8

= [(x + 2)(x + 5)][(x + 3)(x + 4)] - 8

= (x² + 5x + 2x + 10)(x² + 4x + 3x + 12) - 8

= (x² + 7x + 10)(x² + 7x + 12) - 8

Thay x² + 7x + 10 = y, ta có:

y(y + 2) - 8

= y² + 2y - 8

= y² - 2y + 4y - 8

= (y² - 2y) + (4y - 8)

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

= (y + 4)(y - 2)

= (x² + 7x + 10 + 4)(x² + 7x + 10 - 2)

= (x² + 7x + 14)(x² + 7x + 8)

c) x³ - x²y + 3x - 3y

= (x³ - x²y) + (3x - 3y)

= x²(x - y) + 3(x - y)

= (x² + 3)(x - y)

Answer 4

a, x³-xy²+3x-3y

= ( x³-x²y)+(3x-3y)

= x²(x-y)+3(x-y)

=(x-y)(x²+3)

b,x³-2x²-4xy²+x

=x(x²-2x-4y²+1)

=x[(x²-2x+1)-4y²]

=x[(x-1)²-(2y)²]

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

Answer 5