\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)

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

\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)

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

\(\Rightarrow\left(x-5\right)\left(2x+3\right)\)

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

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

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

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

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

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

\(d,x^2-x-\left(4x-4\right)\)

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

\(\Rightarrow x\left(x-1\right)-4\left(x-1\right)\)

\(\Rightarrow\left(x-1\right)\left(x-4\right)\)

\(e,\left(2x-5\right)^2-\left(x+2\right)^2\)

\(\Rightarrow\left(2x-5-x-2\right)\left(2x-5+x+2\right)\)

\(\Rightarrow\left(x-7\right)\left(3x-3\right)\)

\(f,x^2+7x-8\)

\(\Rightarrow x^2-x+8x-8\)

\(\Rightarrow x\left(x-1\right)+8\left(x-1\right)\)

\(\Rightarrow\left(x-1\right)\left(x+8\right)\)

\(a,=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\\ b,=\dfrac{1}{3}x\left(y+3xz+3z\right)\\ c,=2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)

\(d,=x^2\left(\dfrac{2}{5}+5x+y\right)\\ e,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ f,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ g,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)

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) x² + 6x + 8

= x² + 2x + 4x + 8

= (x² + 2x) + (4x + 8)

= x(x + 2) + 4(x + 8)

= (x + 2)(x + 4)

b) 3x² - 2(x - y)² - 3y²

= (3x² - 3y²) - 2(x - y)²

= 3(x² - y²) - 2(x - y)²

= 3(x + y)(x - y) - 2(x - y)²

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

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

= (x - y)(x + 5y)

c) 4x² - 9y² + 4x - 6y

= (4x² - 9y²) + (4x - 6y)

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

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

d) x(x + 1)² + x(x - 5) - 5(x + 1)²

= [x(x + 1)² - 5(x + 1)²] + x(x - 5)

= (x + 1)²(x - 5) + x(x - 5)

= (x - 5)[(x + 1)² + x]

= (x - 5)(x² + 2x + 1 + x)

= (x - 5)(x² + 3x + 1)

e) 2xy - x² + 3y² - 4y + 1

= -x² + 2xy - y² + 4y² - 4y + 1

= -(x² - 2xy + y²) + (4y² - 4y + 1)

= -(x - y)² + (2y - 1)²

= (2y - 1)² - (x - y)²

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

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

f) 4x¹⁶ + 81

= (2x⁸)² + 2.2x⁸.9 + 9² - 2.2x⁸.9

= (2x⁸ + 9)² - 36x⁸

= (2x⁸ + 9) - (6x⁴)²

= (2x⁸ + 9 - 6x⁴)(2x⁸ + 9 + 6x⁴)

= (2x⁸ - 6x⁴ + 9)(2x⁸ + 6x⁴ + 9)

b) \(3x^2+2x-5=3\left(x-1\right)\left(x+\dfrac{5}{3}\right)\)

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

d) \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

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

b: \(3x^2+2x-5\)

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

\(=\left(x-1\right)\left(3x+5\right)\)

c: \(3-2x-x^2\)

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

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

d: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

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

Đáp án A.

Chọn C.

Đặt  u   =   G ( x ) d v   =   f ( x ) d x ⇒ d u   =   G ( x ) ' d x   =   g ( x )   d x v   =   ∫ f ( x ) d x   =   F ( x )

Suy ra: I =  G ( x ) F ( x ) 2 0   - ∫ 0 2 F ( x ) g ( x ) d x  

= G(2)F(2) – G(0)F(0) – 3 = 1 – 0 – 3 = -2.