a: \(16x^3+0,25yz^3\)

\(=0,25\cdot x^3\cdot64+0,25\cdot yz^3\)

\(=0,25\left(64x^3+yz^3\right)\)

b: \(x^4-4x^3+4x^2\)

\(=x^2\cdot x^2-x^2\cdot4x+x^2\cdot4\)

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

c: \(x^3+x^2y-xy^2-y^3\)

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

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

\(=\left(x+y\right)\left(x-y\right)\left(x+y\right)\)

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

d: \(x^3+x^2+x+1\)

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

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

e: \(x^4-x^2+2x-1\)

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

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

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

f: \(2x^2-18\)

\(=2\cdot x^2-2\cdot9\)

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

g: \(x^2+8x+7\)

\(=x^2+x+7x+7\)

\(=x\left(x+1\right)+7\cdot\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)

h: \(x^4y^4+4\)

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

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

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

i: \(x^4+4y^4\)

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

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

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

k: \(x^2-2x-15\)

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

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

a: =(6x)^2-(3x-2)^2

=(6x-3x+2)(6x+3x-2)

=(9x-2)(3x+2)

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

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

=8x(x^2+1)

e: =(4x)^2-2*4x*3y+(3y)^2

=(4x-3y)^2

f: \(=-\left(\dfrac{1}{4}x^4-2\cdot\dfrac{1}{2}x^2\cdot2y^3+4y^6\right)\)

\(=-\left(\dfrac{1}{2}x^2-2y^3\right)^2\)

g: =(4x)^3+1^3

=(4x+1)(16x^2-4x+1)

k: =x^3(27x^3-8)

=x^3(3x-2)(9x^2+6x+4)

l: =(x^3-y^3)(x^3+y^3)

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

$36x^2-(3x-2)^2$

$=[6x-(3x-2)][6x+(3x-2)]$

$=(6x-3x+2)(6x+3x-2)$

$=(3x+2)(9x-2)$

$16(4x+5)^5-25(2x+2)^2$

Câu này không có dạng hằng đẳng thức hiệu hai bình phương vì số mũ của hai phần là $5$ và $2$.

Có thể đặt nhân tử chung $1$ nhưng không phân tích tiếp được bằng các phương pháp thông thường.

$(x-y+4)^2$

Đây đã là một tích:

$(x-y+4)^2=(x-y+4)(x-y+4)$

$(x+1)^4-(x-1)^4$

$=[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]$

$=[x^2+2x+1-x^2+2x-1][x^2+2x+1+x^2-2x+1]$

$=4x(2x^2+2)$

$=8x(x^2+1)$

$16x^2-24xy+9y^2$

$=(4x)^2-2\cdot4x\cdot3y+(3y)^2$

$=(4x-3y)^2$

$-\dfrac{x^4}{4}+2x^2y^3-4y^6$

$=-\dfrac14(x^4-8x^2y^3+16y^6)$

$=-\dfrac14(x^2-4y^3)^2$

$64x^3+1$

$=(4x)^3+1^3$

$=(4x+1)(16x^2-4x+1)$

$x^3y^6z^9-125$

$=(xy^2z^3)^3-5^3$

$=(xy^2z^3-5)(x^2y^4z^6+5xy^2z^3+25)$

$27x^6-8x^3$

$=x^3(27x^3-8)$

$=x^3[(3x)^3-2^3]$

$=x^3(3x-2)(9x^2+6x+4)$

$x^6-y^6$

$=(x^3-y^3)(x^3+y^3)$

$=(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)$

$27x^3-54x^2y+36xy^2-8y^3$

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

$=(3x-2y)^3$

$y^9-9x^2y^6+27x^4y^3-27x^6$

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

$=(y^3-3x^2)^3$

Câu 1: A

Câu 21: A

\(16,A\\ 17,C\\ 18,A\\ 19,C\\ 20,A\\ 21,A\)

a: \(4x^2-12x+9-3\left(2x-3\right)\left(x+1\right)\)

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

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

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

b: \(25-4x^2+8xy-4y^2\)

\(=25-\left(2x-2y\right)^2\)

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

a 4x -4y +(x-y)^2

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

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

c x^2(x+1)-4(x+1)

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

d x^4-(x^2-2x+1)

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

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

MIK KO BIT DUNG HAY KO CON B THI MIK KO BIET LAM

Câu b dễ thôi

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

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

\(=x\left(x+2\right)\left(x^2-6x+4\right)\)

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

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

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

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

ỳtct7ct7c7c7t79tc9

\(a,\Rightarrow\left(x-3-5+2x\right)\left(x-3+5-2x\right)=0\\ \Rightarrow\left(3x-8\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{8}{3}\end{matrix}\right.\\ b,=\left(x+y\right)^2-\left(x-2y\right)^2\\ =\left(x+y-x+2y\right)\left(x+y+x-2y\right)=3y\left(2x-y\right)\\ c,=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)\\ d,=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

\(a/\)

\(4x-4y+x^2-2xy+y^2\)

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

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

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

\(b/\)

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

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

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

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

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

\(=x\left(x+2\right)\left(x^2-6x-4\right)\)

\(d/\)

\(x^4-x^2+2x-1\)

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

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

\(e/\)(Xem lại đề)

\(x^4+x^3+x^2+2x+1\)

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

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

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

\(f/\)

\(x^3-4x^2+4x-1\)

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

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

\(=[\sqrt{x}\left(x-2\right)]^2-1\)

\(=[\sqrt{x}\left(x-2\right)-1][\sqrt{x}\left(x-2\right)+1]\)

\(c/\)

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

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

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

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

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

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

1 ) ( 2x - 1 ) ( 8x + 12 ) + x²( 2x - 1 ) + ( 1 - 2x ).( 2x - 3 )

= ( 2x - 1 ) . ( 8x + 12 ) + x² ( 2x - 1 ) - ( 2x - 1 ) . ( 2x - 3 )

= ( 2x - 1 ) . ( 8x + 12 + x² - 2x + 3 )

= ( 2x - 1 ) . ( x² + 6x + 15 )

2 ) 3x ( x - y ) - 2y ( y - x ) - 4x + 4y

= 3x ( x - y ) + 2y ( x - y ) - 4. ( x - y )

= ( x - y ) ( 3x + 2y  - 4 )