12x^3+16x^2-5x-3 = 12x³+18x² -2x²-3x -2x -3

= 12x²(x +3/2) -2x(x+3/2) -2(x+3/2)

=(12x²-2x-2)(x+3/2)

=2(6x²-x-1)(x+3/2)

=2[6x(x+1/3)-3(x+1/3)](x+3/4)

=2(6x-3)(x+1/3)(x+3/4)

=6(2x-1)(x+1/3)(x+3/4)

k cho mk  nha

 = (12x^3-6x^2)+(22x^2-11x)+(6x-3)

 = (2x-1).(6x^2+11x+3)

 = (2x-1).[(6x^2+2x)+(9x+3)]

 = (2x-1).(3x+1).(2x+3)

k mk nha

 Ta có: \(12x^3+16x^2-5x-3\)

\(=12x^3-6x^2+22x^2-11x+6x-3\)

\(=6x^2\left(2x-1\right)+11x\left(2x-1\right)+3\left(2x-1\right)\)

\(=\left(2x-1\right)\left(6x^2+11x+3\right)\)

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

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

`@` `\text {Ans}`

`\downarrow`

`x^2 + 4x + 3`

`= x^2 + 3x + x + 3`

`= (x^2 + 3x) + (x + 3)`

`= x(x + 3) + (x + 3)`

`= (x+1)(x+3)`

____

`2x^2 + 3x - 5`

`= 2x^2 + 5x - 2x - 5`

`= (2x^2 - 2x) + (5x - 5)`

`= 2x(x - 1) + 5(x - 1)`

`= (2x + 5)(x - 1)`

____

`16x - 5x^2 - 3`

`= 15x + x - 5x^2 - 3`

`= (15x - 5x^2) + (x - 3)`

`= 5x(3 - x) + (x - 3)`

`= -5x(x - 3) + (x - 3)`

`= (1 - 5x)(x - 3)`

\(x^2+4x+3=x^2+3x+x+3=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\\----\\ 2x^2+3x-5=2x^2-2x+5x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\\ ----\\ 16x-5x^2-3=-5x^2+15x+x-3=-5x\left(x+3\right)-\left(x+3\right)=-\left(5x+1\right)\left(x+3\right)\)

\(x^2+4x+3=\left(x+1\right)\left(x+3\right)\\ 2x^2+3x-5=\left(2x+5\right)\left(x-1\right)\\ 16x-5x^2-3=\left(1-5x\right)\left(x-3\right)\)

a) Ta có: \(4x^2-28xy+49y^2\)

\(=\left(2x\right)^2-2\cdot2x\cdot7y+\left(7y\right)^2\)

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

b) Ta có: \(x^2+8xy+16y^2\)

\(=x^2+2\cdot x\cdot4y+\left(4y\right)^2\)

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

c) Ta có: \(x^2-12x+36\)

\(=x^2-2\cdot x\cdot6+6^2\)

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

\(\left(2x-7y\right)^2\)

\(\left(6-x\right)^2\)

4x² - 28xy + 49y²

= (2x)² - 2.2x.7y + (7y)²

= (2x - 7y)²

x² + 8x²y + 16x²y² đề có bị thiếu không , nên mình bổ sung nhé

= x² + 2.x.4xy + (4xy)²

= (x + 4xy)²

36 - 12x + x²

= 6² + 2.6.x + x²

= (6 + x)²

 Chúc bạn học tốt

7,      4x mũ 2 - 12x + 9 - y mũ 2 =  -(y-2x+3) (y+2x-3)

8,      16x mũ 2 - 4y mũ 2 + 4y - 1 =   -(2y - 4x - 1) (2y+4x-1)

9,        25 - x mũ 2 - 12x - 36 =  -(x+1) (x+11)

10,        x mũ 2 - 9 - 5 ( x + 3 ) =  (x-8) (x+3)

bạn k cho mình nha 

chúc bạn học tốt :))))

bạn kham khảo link, mình đã làm rồi nhé

Câu hỏi của Phạm Đỗ Bảo Ngọc - Toán lớp 8 - Học trực tuyến OLM 

Trả lời:

7, 4x² - 12x + 9 - y²

= ( 4x² - 12x + 9 ) - y² 

= ( 2x - 3 )² - y² 

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

8, 16x² - 4y² + 4y - 1

= 16x² - ( 4y² - 4y + 1 )

= ( 4x )² - ( 2y - 1 )²

= ( 4x - 2y + 1 )( 4x + 2y - 1 )

9, 25 - x² - 12x - 36

= 25 - ( x² + 12x + 36 )

= 5² - ( x + 6 )²

= ( 5 - x - 6 )( 5 + x + 6 )

= ( - 1 - x )( 11 + x )

10, x² - 9 - 5 ( x + 3 )

= ( x - 3 )( x + 3 ) - 5 ( x + 3 )

= ( x + 3 )( x - 3 - 5 )

= ( x + 3 )( x - 8 )

đa thức này đã phân tích thành nhân tử rồi bạn ạ

1) \(\left(3x^2-3y^2\right)-\left(12x-12y\right)\)

\(=3xy\left(x-y\right)-12\left(x-y\right)\)

\(=\left(3xy-12\right)\left(x-y\right)\)

2) \(4x^3+4xy^2+8x^2y-16x\)

\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)

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

Ta có : 3x² - 3y² - 12x + 12y 

= (3x² - 3y²) - (12x - 12y)

= 3(x² - y²) - 12(x - y)

= 3(x - y)(x + y) - 4.3.(x - y)

= 3(x - y)(x + y - 4)

3) x^4-5x^2+4 = x^4-x^2-4x^2+4 

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

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

2:

a: \(x^2-12x+20\)

\(=x^2-2x-10x+20\)

=x(x-2)-10(x-2)

=(x-2)(x-10)

b: \(2x^2-x-15\)

=2x^2-6x+5x-15

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

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

c: \(x^3-x^2+x-1\)

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

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

d: \(2x^3-5x-6\)

\(=2x^3-4x^2+4x^2-8x+3x-6\)

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

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

e: \(4y^4+1\)

\(=4y^4+4y^2+1-4y^2\)

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

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

f; \(x^7+x^5+x^3\)

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

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

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

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

g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)

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

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

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

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

h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)

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

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

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

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

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

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

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

i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)

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

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

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

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

j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)

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

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

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