đề bài của bài này là tính thuii ạ

a) \(x^3+3x^2+3x+1=x^2+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x-1\right)^3\)

b) \(x^2+6x+9=x^2+2\cdot3\cdot x+3^2=\left(x+3\right)^2\)

c) \(-x^3+9x^2-27x+27\)

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

\(=-\left(x^3-3\cdot3\cdot x^2+3\cdot3^2\cdot x-3^3\right)=-\left(x-3\right)^3\)

d) \(x^2+4x+4=x^2+2\cdot2\cdot x+2^2=\left(x+2\right)^2\)

k) \(10x-25-x^2=-x^2+10x-25=-\left(x^2-10x+25\right)\)

\(=-\left(x^2-2\cdot5\cdot x+5^2\right)=-\left(x-5\right)^2\)

f) \(\left(x+y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left[\left(x-y\right)-3x\right]\left[\left(x-y\right)+3x\right]\)

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

a: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)

b: \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)

c: \(2x-1-x^2\)

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

d: \(x^2+x+\dfrac{1}{4}=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

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

g: \(\left(x+5\right)^2-4x^2=\left(x+5+2x\right)\left(x+5-2x\right)\)

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

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

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

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

i: \(=x^2y^2-4xy+4-3\)

\(=\left(xy-2\right)^2-3=\left(xy-2-\sqrt{3}\right)\left(xy-2+\sqrt{3}\right)\)

k: \(=y^2-\left(x-1\right)^2\)

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

l: \(=x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=\left(x+2\right)^3\)

m: \(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2-y^3=\left(2x-y\right)^3\)

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

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

d.

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

e.

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

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

f.

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

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2

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

\(=\left(2x+y\right).3y\)

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

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

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

c) \(9x^2-3x+2y-4y^2\)

\(=9x^2-4y^2-3x+2y\)

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

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

d) \(4x^2-4xy+2x-y+y^2\)

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

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

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

e) \(x^3+3x^2+3x+1-y^3\)

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

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

g) \(x^3-2x^2y+xy^2-4x\)

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

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

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

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

a) (x + 2y)² - (x - y)²

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

= 3y(2x + y)

b) (x + 1)³ + (x - 1)³

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

= 2x(x² + 2x + 1 - x² + 1 + x² - 2x + 1)

= 2x(x² + 3)

c) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

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

= (3x - 2y)(3x + 2y - 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

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

e) x³ + 3x² + 3x + 1 - y³

= (x³ + 3x² + 3x + 1) - y³

= (x + 1)³ - y³

= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]

= (x - y + 1)(x² + 2x + 1 + xy + y + y²)

g) x³ - 2x²y + xy² - 4x

= x(x² - 2xy + y² - 4)

= x[(x² - 2xy + y²) - 4]

= x[(x - y)² - 2²]

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

a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

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

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

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

b) \(\left(6x^5y^2-9x^4y^3+15x^3y^4\right):3x^3y^2\)

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

\(a,=8\left(x^3-125\right)=8\left(x-5\right)\left(x^2+5x+25\right)\\ b,=\left(0,1+4x\right)\left(0,01-0,4x+16x^2\right)\\ c,=\left(x+\dfrac{1}{5}y\right)\left(x^2-\dfrac{1}{5}xy+\dfrac{1}{25}y^2\right)\\ d,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ e,=\left(x-1+3\right)\left[\left(x-1\right)^2-3\left(x-1\right)+9\right]\\ =\left(x+2\right)\left(x^2-2x+1-3x+3+9\right)\\ =\left(x+2\right)\left(x^2-5x+13\right)\\ f,=\left(\dfrac{x^2}{2}-y^2\right)\left(\dfrac{x^4}{4}+\dfrac{x^2y^2}{2}+y^4\right)\)