Bài 3

a) x² + 10x + 25

= x² + 2.x.5 + 5²

= (x + 5)²

b) 8x - 16 - x²

= -(x² - 8x + 16)

= -(x² - 2.x.4 + 4²)

= -(x - 4)²

c) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³

d) (x + y)² - 9x²

= (x + y)² - (3x)²

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

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

e) (x + 5)² - (2x - 1)²

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

= (6 - x)(3x + 4)

Bài 4

a) x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

b) (x - 4)² - 36 = 0

(x - 4 - 6)(x - 4 + 6) = 0

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

x - 10 = 0 hoặc x + 2 = 0

*) x - 10 = 0

x = 10

*) x + 2 = 0

x = -2

Vậy x = -2; x = 10

c) x² - 10x = -25

x² - 10x + 25 = 0

(x - 5)² = 0

x - 5 = 0

x = 5

d) x² + 5x + 6 = 0

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 = -3; x = -2

từng câu 1 thôi:v

a) x2-xy+5y-25
 = x(2-y)+ 5(y-2)
 = x(2-y)-5(2-y)
 = (x-5)(2-y)

h: \(=\left(a-1-b\right)\left(a-1+b\right)\)

a) \(=x^4-14x^2+40-72=x^4-14x^2-32=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)

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

c) \(=x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5=x^4+6x^3+7x^2-6x-8=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

a: Ta có: \(\left(x^2-4\right)\left(x^2-10\right)-72\)

\(=x^4-14x^2-32\)

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

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

b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left(x^2+5x+6\right)\left(x^2+5x+4\right)+1\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)

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

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)

\(a,=2x\left(x+3\right)\\ b,=x^3\left(x+3\right)+\left(x+3\right)=\left(x^3+1\right)\left(x+3\right)\\ =\left(x+1\right)\left(x+3\right)\left(x^2-x+1\right)\\ c,=64-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\\ A=x^2+6x+5+x^3-8-x^2-x+2\\ A=x^3+5x-1\)

a) 2x²+6x=2x(x+3)
b) x⁴+3x³+x+3=(x⁴+x)+(3x³+3)=x(x³+1)+3(x³+1)=(x+3)(x³+1)
c) 64-x²-y²+2xy=-(x²-2xy+y²)+8²=8-(x+y)²=(8+x+y)(8-x-y)

A= (x+5)(x+1)+(x-2)(x²+2xx+4)-(x²+x-2)
A= x²+6x+5+x³-8-x²-x+2
A= x³+(x²-x²)+(6x-x)+(5-8+2)
A= x³+5x-1

\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Câu 1

a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)

b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)

Bài 2

a) \(7x^2+14xy=7x\left(x+2y\right)\)

b) \(3x+12-\left(x^2+4x\right)=-x^2-x+12=\left(-x+3\right)\left(x+4\right)\)

c) \(x^2-2xy+y^2=\left(x-y\right)^2\)

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

a: =3x(y-2)-5(y-2)

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

b: =(2x-5y)(2x+5y)

c: =x(x^3-8)

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

a)

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

b)

\(4x^2-25y^2\\ =\left(2x-5y\right)\left(2x+5y\right)\)

c)

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

\(a,3x\left(y-2\right)+5\left(2-y\right)=3x\left(y-2\right)-5\left(y-2\right)=\left(y-2\right)\left(3x-5\right)\)

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

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

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

b) \(\left(1+x^2\right)\left(1+y^2\right)+4xy+2\left(x+y\right)\left(1+xy\right)=25\Leftrightarrow1+x^2+y^2+x^2y^2+4xy+2\left(x+y\right)\left(1+xy\right)-25=0\Leftrightarrow\left(x+y\right)^2+2\left(x+y\right)\left(1+xy\right)+\left(1+xy\right)^2-25=0\Leftrightarrow\left(x+y+1+xy\right)^2-25=0\Leftrightarrow\left(x+y+xy-24\right)\left(x+y+xy+26\right)=0\)

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

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

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

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