a) \(x^3+4x^2-21x\)

\(=x\left(x^2+4x-21\right)\)

\(=x\left(x^2-3x+7x-21\right)\)

\(=x\left[x\left(x-3\right)+7\left(x-3\right)\right]\)

\(=x\left(x-3\right)\left(x+7\right)\)

b) \(5x^3+6x^2+x\)

\(=x\left(5x^2+6x+1\right)\)

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

\(=x\left[5x\left(x+1\right)+\left(x+1\right)\right]\)

\(=x\left(x+1\right)\left(5x+1\right)\)

c) \(x^3-7x+6\)

\(=x^3+2x^2-3x-2x^2-4x+6\)

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

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

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

d) \(3x^3+2x-5\)

\(=3x^3+3x^2+5x-3x^2-3x-5\)

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

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

a) x^2 + 2x + 1

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

\(=\left(x+1\right)^2\)
b) x^2 + 6x + 9

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

\(=\left(x+3\right)^2\)
c) x^2 - 6x + 9

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

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

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

\(=\left(x-1\right)^2\)
e ) 4x^2 + 4xy +y^2

\(=\left(2x\right)^2+2.2x.y+y^2\)

\(=\left(2x+y\right)^2\)
f) x^2 + 4xy + 4y^2

\(=x^2+2.x.2y+\left(2y\right)^2\)

\(=\left(x+2y\right)^2\)

phân tích các đa thức sau thành nhân tử
a) x^2 + 2x + 1

\(=\left(x+1\right)^2\)
b) x^2 + 6x + 9

\(=\left(x+3\right)^2\)
c) x^2 - 6x + 9

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

\(=\left(x-1\right)^2\)
e ) 4x^2 + 4xy +y^2

\(=\left(2x\right)^2+4xy+y^2\)

\(=\left(2x+y\right)^2\)
f) x^2 + 4xy + 4y^2

\(=x^2+4xy+\left(2y\right)^2\)

\(=\left(x+2y\right)^2\)

p/s: --.--

bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được

a) Áp dụng HĐT 1 thu được ( 2 x   +   y ) 2 .

b) Áp dụng HĐT 3 với A = 2x + l; B = x - l thu được

[(2x +1) + (x -1)] [(2x +1) - (x -1)] rút gọn thành 3x(x + 2).

c) Ta có: 9 - 6x +  x 2  -  y 2 = ( 3   -   x ) 2  -  y 2  = (3 - x - y)(3 -x + y).

d) Ta có: -(x + 2) + 3( x 2  - 4) = -{x + 2) + 3(x + 2)(x - 2)

= (x + 2) [-1 + 3(x - 2)] = (x + 2)(3x - 7).

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

1

a, 2x²+4x+2-2y² = 2(x²+2x+1-y²)= 2[(x+1)²-y² ] = 2(x-y+1)(x+y+1)

b, 2x - 2y - x² + 2xy - y²= 2(x -y) - (x² - 2xy + y²) = 2(x-y)-(x-y)²=(x-y)(2-x+y)

c, x²-y²-2y-1=x²-(y²+2y+1)=x²-(y+1)²=(x-y-1)(x+y+1)

d, x²-4x-2xy-4y+y²= x²-2xy+y²-4x-4y=(x-y)

2.

a, x²-3x+2=x²-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)

b, x²+5x+6=x²+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)

c, x²+6x-6=

a. \(=4x^3-12x^2-x^2+3x+6x-18=\left(x-3\right)\left(4x^2-x+6\right)\)

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

c.  \(=x^3+2x^2-6x^2-12x+4x+8=\left(x+2\right)\left(x^2-6x+4\right)\)

a) \(xy+y^2-x-y=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)

b) \(25-x^2+4xy-4y^2=25-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)

c) \(x^2-4x+3=x^2-x-3x+3=x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(x-3\right)\)

d) \(y^2\left(x-1\right)-7y^3+7xy^3\)

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

\(=y^2\left[\left(x-1\right)-7y\left(1-x\right)\right]=y^2\left(x-1\right)\left(1+7y\right)\)

a)

 \(xy+y^2-x-y\\ =\left(xy-x\right)+\left(y^2-y\right)\\ =x\left(y-1\right)+y\left(y-1\right)\\ =\left(y-1\right)\left(x+y\right)\)

b)

 \(25-x^2+4xy-4y^2\\ =25-\left(x^2-4xy-4y^2\right)\\ =5^5-\left(x-y\right)^2\\ =\left(5+x-y\right)\left(5-x+y\right)\)

a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)

b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)

c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)

d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)

tik nhé

​