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)

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 ) 

a) (x³-x²)+(8x-8)=x(x-1)+8(x-1)=(x²+8)(x-1)

b) 8x³-8x²y+2xy²=2x(4x²-4xy+y²)

c) (x²+y²-z²)² - 4x²y²=(x²+y²-z²)² - (2xy)²=(x²+y²-z²-2xy)(x²+y²-z²+2xy)

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

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

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

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

5, \(x^4+8x=x\left(x^3+8\right)=x\left(x+8\right)\left(x^2-8x+64\right)\)

\(1,\)

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

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

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

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

\(2,\)

\(x^2-25+y^2+2xy\)

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

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

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

\(3,\)

\(x^2y-x^3-9y+9x\)

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

\(=x^2\left(y-x\right)-9\left(y-x\right)\)

\(=\left(x^2-9\right)\left(y-x\right)\)

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

\(4,\)

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

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

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

\(5,\)

\(x^4-8x\)

\(=x\left(x^3-8\right)\)

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

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Bài 1

a) 5x²y - 20xy²

= 5xy(x - 4y)

b) 1 - 8x + 16x² - y²

= (1 - 8x + 16x²) - y²

= (1 - 4x)² - y²

= (1 - 4x - y)(1 - 4x + y)

c) 4x - 4 - x²

= -(x² - 4x + 4)

= -(x - 2)²

d) x³ - 2x² + x - xy²

= x(x² - 2x + 1 - y²)

= x[(x² - 2x+ 1) - y²]

= x[(x - 1)² - y²]

= x(x - 1 - y)(x - 1 + y)

= x(x - y - 1)(x + y - 1)

e) 27 - 3x²

= 3(9 - x²)

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

f) 2x² + 4x + 2 - 2y²

= 2(x² + 2x + 1 - y²)

= 2[(x² + 2x + 1) - y²]

= 2[(x + 1)² - y²]

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

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

Bài 2:

a: \(x^2\left(x-2023\right)+x-2023=0\)

=>\(\left(x-2023\right)\left(x^2+1\right)=0\)

mà \(x^2+1>=1>0\forall x\)

nên x-2023=0

=>x=2023

b: 

ĐKXĐ: x<>0

\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)

=>\(-x\left(x-4\right)+2x^2-4x-9=0\)

=>\(-x^2+4x+2x^2-4x-9=0\)

=>\(x^2-9=0\)

=>(x-3)(x+3)=0

=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

c: \(x^2+2x-3x-6=0\)

=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)

=>\(x\left(x+2\right)-3\left(x+2\right)=0\)

=>(x+2)(x-3)=0

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

d: 3x(x-10)-2x+20=0

=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)

=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)

=>\(\left(x-10\right)\left(3x-2\right)=0\)

=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)

Câu 1:

a: \(5x^2y-20xy^2\)

\(=5xy\cdot x-5xy\cdot4y\)

\(=5xy\left(x-4y\right)\)

b: \(1-8x+16x^2-y^2\)

\(=\left(16x^2-8x+1\right)-y^2\)

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

\(=\left(4x-1-y\right)\left(4x-1+y\right)\)

c: \(4x-4-x^2\)

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

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

d: \(x^3-2x^2+x-xy^2\)

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

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

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

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

e: \(27-3x^2\)

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

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

f: \(2x^2+4x+2-2y^2\)

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

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

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

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

Bài 2

a) x²(x - 2023) - 2023 + x = 0

x²(x - 2023) - (x - 2023) = 0

(x - 2023)(x² - 1) = 0

x - 2023 = 0 hoặc x² - 1 = 0

*) x - 2023 = 0

x = 2023

*) x² - 1 = 0

x² = 1

x = 1 hoặc x = -1

Vậy x = -1; x = 1; x = 2023

b) -x(x - 4) + (2x³ - 4x² - 9x) : x = 0

-x² + 4x + 2x² - 4x - 9 = 0

x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

Vậy x = 3; x = -3

c) x² + 2x - 3x - 6 = 0

(x² + 2x) - (3x + 6) = 0

x(x + 2) - 3(x + 2) = 0

(x + 2)(x - 3) = 0

x + 2 = 0 hoặc x - 3 = 0

*) x + 2 = 0

x = -2

*) x - 3 = 0

x = 3

Vậy x = -2; x = 3

d) 3x(x - 10) - 2x + 20 = 0

3x(x - 10) - (2x - 20) = 0

3x(x - 10) - 2(x - 10) = 0

(x - 10)(3x - 2) = 0

x - 10 = 0 hoặc 3x - 2 = 0

*) x - 10 = 0

x = 10

*) 3x - 2 = 0

3x = 2

x = 2/3

Vậy x = 2/3; x = 10

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

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

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

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

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

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

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

\(10x\left(x-y\right)-6y\left(y-x\right)\)

\(=10x\left(x-y\right)+6x\left(x-y\right)\)

\(=\left(10x+6x\right)\left(x-y\right)\)

\(c,3x^2+5y-3xy-5x\)

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

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

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

\(e,27+27x+9x^2=3\left(9+9x+x^2\right)\)

\(f,8x^3-12x^2y+6xy^2-y^3\)

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

\(g,x^3+8y^3=x^3+\left(2y\right)^3\)

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

\(i,x^2-25-2xy+y^2\)

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

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