x³-2x²-5x=x(x²-2x-5)

\(x^3-2x^2-5x=x\left(x^2-2x-5\right)\)

x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)

k mk nha

bạn ơi bạn chưa bớt 2x^2 kìa

ngu dốt

x⁵-x⁴-1=x⁵-x³-x²-x⁴+x²+x+x³-x-1

=x².(x³-x-1)-x.(x³-x-1)+(x³-x-1)

=(x³-x-1)(x²-x+1)

x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)

k mk nha

mik chịu

= (-2a ^2 +4a)(1-x)

= -2 (a^2 -2a)(1-x)

= -2a(a-2)(1-x)

= 2a(a-2)(x-1)

\(-2a^2\left(x-1\right)+4a\left(1-x\right)\)

\(=-a\cdot2a\left(x-1\right)-2\cdot2a\left(x-1\right)\)

\(=2a\left(x-1\right)\left(-a-2\right)\)

\(7xy^5\left(x-1\right)-3x^2y^4\left(1-x\right)+5xy^3\left(x-1\right)\)

\(=7xy^5\left(x-1\right)+3x^2y^4\left(x-1\right)+6xy^3\left(x-1\right)\)

\(=\left(x-1\right)\left(7xy^5+3x^2y^4-6xy^3\right)=xy\left(x-1\right)\left(7y^4+3xy^3-6y^2\right)\)

 Trả lời:

7xy⁵(x - 1) - 3x²y⁴(1 - x) + 5xy³(x - 1)

= 7xy⁵(x - 1) + 3x²y⁴(x - 1) + 5xy³(x - 1)

= (7xy⁵ + 3x²y⁴ + 5xy³)(x - 1)

= xy(7y⁴ + 3xy³ + 5y²)(x - 1)

\(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)

\(=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)\)

\(=\left(x+1\right)\left(-2x^2+3x+7\right)\)

a) x.(x-1) - 2(1-x) =0

=> x.(x-1) + 2(x-1) =0

=> (x-1)(x+2) =0

b) (x-3)^3 + (3-x) =0

=> (x-3)^3 - (x-3) =0

=> (x-3)​​[(x-3)^2 - 1 ] =0

=> (x-3)(x-3-1)(x-3+1) =0

=> (x-3)(x-4)(x-2) =0

x² + x -12 = x² + 4x - 3x - 12 = x(x+4) - 3(x+4) = (x+4)(x-3)

\(x^2+x-12\)

\(=x^2+x+\frac{1}{4}-\frac{49}{4}\)

\(=\left(x+\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)

\(=\left(x+\frac{1}{2}-\frac{7}{2}\right)\left(x+\frac{1}{2}+\frac{7}{2}\right)\)

\(=\left(x-3\right)\left(x+4\right)\)

\(3,\)Nhẩm nghiệm của đa thức trên ta đc : -1

Ta có lược đồ sau :

Phân tích thành nhân tử ta có :\(\left(x+1\right)\left(x^2-4\right)\)