\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)

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

b: =x^2-6xy+9y^2=(x-3y)^2

c: =5x(x^2-2xy+y^2)

=5x(x-y)^2

d: =(x-3)^2

e: =(2y-z)(4x+7y)

a)HĐT:(x^2+1-x)(x^2+1+x)

b)=x^2-2.x.3y+(3y)^2

c)=5x(x^2-2xy+y^2)

=5x(x-y)^2

d)x^2-2.3.x+3^2

=(x-3)^2

e)(2y-z)+7y(2y-z)

=(2y-z)(1+7y)

\(\left(x-1\right)^2-25\)

\(=x^2-2x+1-25\)

\(=x^2-2x-24\)

\(=x^2-6x+4x-24\)

\(=x.\left(x-6\right)+4.\left(x-6\right)\)

\(=\left(x+4\right).\left(x-6\right)\)

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

b, \(\left(x+1\right)^2-25=\left(x+1\right)^2-5^2=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)

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

d,  \(8-27x^3=2^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

a)=(1-y)²

b)=(x+1)²-5²

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

=(x+6)(x-4)

c)=1²-(2x)²

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

d)=2³-(3x)³

=(2-3x)(4+6x+9x²)

e)=3³+3.9.x+3.3.x²+x³

=(3+x)³

Bài làm:

a, 1-4x²

=1-(2x)²

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

b, 8-27x³

=2³-(3x)³

=(2-3x).(4+6x+9x²)

Các câu còn lại bạn dùng hằng đẳng thức là phân tích được ra thôi

1 - 4x^2 

= 1^2 - ( 2x )^2 

= ( 1 - 2x ) ( 1 + 2x ) 

8 - 27x^ 3 

= 2^3 - ( 3x )^3 

= ( 2 - 3x ) [ 2^2 + 2 * 3x + ( 3x )^2 ]

= ( 2 - 3x ) ( 4 + 6x + 9x^2 ) 

= ( 2 - 3x ) ( 9x^2 + 6x + 4 ) 

27 + 27x + 9x^2 + x^3 

= x^3 + 9x^2 + 27x + 27 

= x^3 + 3x^2 + 6x^2 + 18x + 9x + 27 

= x^2 ( x + 3 ) + 6x ( x + 3 ) + 9 ( x + 3 ) 

= ( x + 3 ) ( x^2 + 6x + 9 ) 

= ( x + 3 ) ( x + 3 )^2 

= ( x + 3 )^3 

x^2 + 4x - 5 

= x^2 - x + 5x - 5 

= x ( x - 1 ) + 5 ( x - 1 ) 

= ( x + 1 ) ( x - 5 ) 

\(1,=8xy+14y^2-4xz-7yz\\ 2,=y\left(4x^2-12x+9\right)=y\left(2x-3\right)^2\\ 3,\Leftrightarrow\left(x+3\right)\left(x-2+x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

Câu 1: \(\left(2y-z\right)\left(4x+7y\right)=8xy-4xz+14y^2-7yz\)

câu 2: \(4x^2y-12xy+9y=y\left(4x^2-12x+9\right)\)

câu 3: \(\left(x-2\right)\left(x+3\right)+x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-2+x\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2x-2\right)=0\\ \Leftrightarrow2\left(x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

\(x^4+6x^3+13x^2+12x+4\)

\(=x^4+x^3+5x^3+5x^2+8x^2+8x+4x+4\)

\(=x^3\left(x+1\right)+5x^2\left(x+1\right)+8x\left(x+1\right)+4\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3+5x^2+8x+4\right)\)

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

\(=\left(x+1\right)\left[x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)\right]\)

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

Bài 1: Thực hiện phép tính.

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

Bài 2: Phân tích đa thức thành nhân tử.

a) \(14x^3y^3-7x^2y+21x^2y^5=7x^2y\left(2xy^2-1+3y^4\right)\)

b) \(18x\left(1-x\right)-12y+12xy=18x\left(1-x\right)-12y\left(1-x\right)=6\left(1-x\right)\left(3x-2y\right)\)

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

a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4 
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))