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)

a) 9x⁴+16y⁶-24x²y³

=(3x²)²-2.3x².4y³+(4y³)²

=(3x²-4y³)²

b) 16x²-24xy+9y²

=(4x)²-2.4x.3y+(3y)²

=(4x-3y)²

c) 36x²-(3x-2)²

=(36x-3x+2)(36x+3x-2)

=(33x+2)(39x-2)

d) 27x³+54x²y+36xy²+8y³

=(3x)³+3.(3x)².2y+3.3x.(2y)²+(2y)³

=(3x+2y)³

e) y⁹-9x²y⁶+27x⁴y³-27x⁶

=(y³)³-3.(y³)².3x²+3.y³.(3x²)²-(3x²)³

=(y³-3x²)³

f) 64x³+1

= (4x)³+1³

=(4x+1)[(4x)²-4x.1+1²]

=(4x+1)(16x²-4x+1)

e) 27x⁶-8x³  *sửa đề*

=(3x²)³-(2x)³

=(3x²-2x)[(3x)²+3x².2x+(2x)²]

=(3x²-2x)(9x²+6x³+4x²)

~~~

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

a: \(=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)

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

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

d: =(x+3)^3

e: \(=\left(2x-y\right)^3\)

f: =(x+2y)(x^2-2xy+4y^2)

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

c: \(x^3-125=\left(x-5\right)\left(x^2+5x+25\right)\)

\(\dfrac{1}{8}x^3-64=\left(\dfrac{1}{2}x-4\right)\left(\dfrac{1}{4}x^2+2x+16\right)\)

d: \(=\left(2x+5y\right)^3\)

a: Ta có: \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2\)

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

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

b: Ta có: \(\left(x-3\right)^2-16\)

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

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

c: \(y^2+16y+64=\left(y+8\right)^2\)

Làm bài 1 thôi !! Mấy bài kia tương tự . Tìm nhân tử chung ra .

a) \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)

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

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

c) \(-16+\left(x-3\right)^2=\left(x-3+4\right)\left(x-3-4\right)=x\left(x-7\right)\)

d) \(64+16y+y^2=\left(y+8\right)\left(y+8\right)\)

a)\(x^3+3xy+y^3-1\)

\(=x^3+3x^2y+3xy^2+y^3-1-3x^2y-3xy^2+3xy\)

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

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

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

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

b) Đặt \(B=3x^2+22xy+11x+37y+7y^2+10\)

Giả sử \(B=\left(ax+by+c\right)\left(mx+ny+p\right)\)

\(=amx^2+anxy+apx+bmxy+bny^2+bpy+cmx+cny+cp\)

\(=amx^2+\left(an+bm\right)xy+\left(ap+cm\right)x+bny^2+\left(bp+cn\right)y+cp\)

Ta được hệ: \(\left\{{}\begin{matrix}am=3;an+bm=22\\ap+cm=11;bn=7\\bp+cn=37;cp=10\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=3;b=1\\c=5;m=1\\n=7;p=2\end{matrix}\right.\)

Vậy B phân tích được thành \(\left(3x+y+5\right)\left(x+7y+2\right)\).

\(a,=4x^2+4x+1\\ b,=9-12y+4y^2\\ c,=\dfrac{x^2}{4}-xy+y^2\\ d,=\dfrac{25}{4}-5x+x^2\\ e,=4x^2+32xy+64y^2\\ f,=9x^2-30xy+25y^2\)

a. (2x + 1)² 

= 4x² + 4x + 1

b. (3 - 2y)²

= 9 - 12y + 4y²

- Các câu còn lại bn dung CT: (A + B)² = A² + 2AB + B² và (A - B)² = A² - 2AB + B² để tính tiếp nha, phân số cũng đc tính.)

a) 4x^2 + 4x + 1

b) 4y^2 - 12y + 9

c) x^2/4 - xy + y^2

d) x^2 - 5x + 25/4

e) 4x^2 + 32xy + 64y^2

f) 9x^2 - 30xy + 25y^2