Question

Phân tích các đa thức thành nhân tử

6x - 9 - x²

5x³ - 10x²y + 5xy²

5x² - 10xy + 5y² - 20z²

x² + 5x + 6

a² +9a + 8

x² - x - 6

x² + 6x + 5

x³ - 3x + 2

Answer 1

a) 6x - 9 - x²

= -( x² - 6x + 3²)

= -(x-3)²

b) 5x³ - 10x²y + 5xy²

= 5x( x² - 2xy + y²)

= 5x(x - y)²

c) 5x² - 10xy + 5y² - 20z²

= 5( x² - 2xy + y² - 4z²)

= 5[( x - y)² - ( 2z)²]

= 5( x - y -2z).(x - y +2z)

d) x² + 5x +6

=( x² + 2.2x + 2² )+ (2 +x)

=( x +2)² + ( x +2)

= ( x+2).( x + 2 + 1)

= ( x +2).( x +3)

e) a² + 9a + 8

= a² + a + 8a + 8

= a( a + 1) + 8( a +1)

= ( a +1).( a +8)

g) x² -x -6

= x² - 2² - x - 2

= ( x -2).(x +2) - ( x +2)

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

= ( x+2).( x -3)

f) x² + 6x +5

= x² + x + 5x + 5

= x( x +1) +5( x +1)

=( x +5).( x +1)

h) x³ - 3x +2

= x³ - x - 2x +2

= x( x² -1) - 2( x -1)

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

= ( x -1).( x² +x -2)

Answer 2

Câu 1 : 6x - 9 - x²

\(=-\left(6x+9+x^2\right)\)

\(=-\left(x^2+6x+9\right)\)

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

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

Câu 2 : 5x³ - 10x²y + 5xy²

* Gợi ý : Câu này ta thấy đa thức có thừa số chung là 5x nên ta phân tích đa thức thành nhân tử bằng phương thức đặt thừa số chung.

Giải :

\(5x^3-10xy^2+5xy^2\)

\(=5x\left(x^2-2xy+y^2\right)\)

Answer 3

\(\text{a) }6x-9-x^2\\ =-\left(x^2+6x+9\right)\\ =-\left(x+3\right)^2\) \(\text{b) }5x^3-10x^2y+5xy^2\\ =5x\left(x^2-2xy+y^2\right)\\ =5x\left(x-y\right)^2\)

\(\text{c) }5x^2-10xy+5y^2-20z^2\\ =5\left(x^2-2xy+y^2-4z^2\right)\\ =5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\\ =5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\\ =5\left(x-y-2z\right)\left(x-y+2z\right)\) \(\text{d) }x^2+5x+6\\ =x^2+3x+2x+6\\ =\left(x^2+3x\right)+\left(2x+6\right)\\ =x\left(x+3\right)+2\left(x+3\right)\\ =\left(x+2\right)\left(x+3\right)\)

\(\text{e) }a^2+9a+8\\ =a^2+a+8a+8\\ =\left(a^2+a\right)+\left(8a+8\right)\\ =a\left(a+1\right)+8\left(a+1\right)\\ =\left(a+8\right)\left(a+1\right)\) \(\text{f) }x^2-x-6\\ =x^2-3x+2x-6\\ =\left(x^2-3x\right)+\left(2x-6\right)\\ =x\left(x-3\right)+2\left(x-3\right)\\ =\left(x+2\right)\left(x-3\right)\)

\(\text{g) }x^2+6x+5\\ =x^2+x+5x+5\\ =\left(x^2+x\right)+\left(5x+5\right)\\ =x\left(x+1\right)+5\left(x+1\right)\\ =\left(x+1\right)\left(x+5\right)\)

\(\text{h) }x^3-3x+2\\=x^3-4x+x+2-2x^2+2x^2+2\\ =\left(x^3-2x^2+x\right)-\left(2x^2-4x+2\right)\\ = x\left(x^2-2x+1\right)-2\left(x^2-2x+1\right)\\ =\left(x-2\right)\left(x^2-2x+1\right)\\ =\left(x-2\right)\left(x-1\right)^2\)