\(a,=4\left(x-1\right)^2\\ b,=\left(x-y\right)\left(x+y\right)+3\left(x+y\right)=\left(x+y\right)\left(x-y+3\right)\)

a, 4x² - 8x + 4 = (2x)² - 2.2x.2 + 2 = (2x - 2)²

b, x² - y² + 3x + 3y = (x² - y²) + (3x + 3y) = (x- y). (x + y) + 3.(x + y) = (x+y).(x- y + 3)

Bài 1: 

a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3-2x+5\right)\)

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

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

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

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

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

\(=\left(3x-2\right)\left(3x-6\right)\)

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

Bài 2: 

a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)

\(=\left(a-b\right)\left(2a-4b\right)\)

\(=2\left(a-b\right)\left(a-2b\right)\)

f: Ta có: \(x^2-6xy+9y^2+4x-12y\)

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

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

whatterfice

b) \(16x-5x^2-3=5x\left(3-x\right)-\left(3-x\right)=\left(3-x\right)\left(5x-1\right)\)

c) \(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)

d) \(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)

\(a,=7xy\left(x^2-2xy+y^2\right)=7xy\left(x-y\right)^2\\ b,=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\\ c,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\)

\(a,15a^2b^3+5a^3b^2=5a^2b^2\left(3b+a\right)\\ b,x^2-2x+1-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)

a) 15a²b³+5a³b²=5a²b²(3b+a)
b) x²-2x+x-y²=( x²-y²⁾-(2x+x)
                     =(x-y)(x+y)-x(2-1)
                      =(x-y)(x+y)-x3

a

\(8x^3-\dfrac{1}{125}y^3\\ =\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\\ =\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\\ =\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)

b

\(-x^3+6x^2y-12xy^2+8y^3\\ =-\left(x^3-6x^2y+12xy^2-8y^3\right)\\ =-\left(x^3-3.2y.x^2+3.\left(2y\right)^2.x-\left(2y\right)^3\right)\\ =-\left(x-2y\right)^3\\ =-\left(x-2y\right)\left(x-2y\right)\left(x-2y\right)\)

a: 8x^3-1/125y^3

=(2x)^3-(1/5y)^3

=(2x-1/5y)(4x^2+2/5xy+1/25y^2)

b: =(2y-x)^3

a) \(8x^3-\dfrac{1}{125}y^3\)

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

\(=\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x\cdot\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\)

\(=\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{24}y^2\right)\)

b) \(-x^3+6x^2y-12xy^2+8y^3\)

\(=-\left(x^3-6x^2y+12xy^2-8y^2\right)\)

\(=-\left(x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\right)\)

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

16ab + 4b² - 9 + 16a²

= (16a² + 16ab + 4b²) - 9

= (4a+2b)² - 3²

= (4a+2b-3)(4a+2b+3)

Lời giải:

a.

\(-16a^4b^6-24a^5b^5-9a^6b^4=-[(4a^2b^3)^2+2.(4a^2b^3).(3a^3b^2)+(3a^3b^2)^2]\)

\(=-(4a^2b^3+3a^3b^2)^2=-[a^2b^2(4b+3a)]^2\)

\(=-a^4b^4(3a+4b)^2\)

b.

$x^3-6x^2y+12xy^2-8x^3$

$=x^3-3.x^2.2y+3.x(2y)^2-(2y)^3=(x-2y)^3$

c.

$x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}$

$=x^3+3.x^2.\frac{1}{2}+3.x.\frac{1}{2^2}+(\frac{1}{2})^3$

$=(x+\frac{1}{2})^3$

a) Ta có: \(-16a^4b^6-24a^5b^5-9a^6b^4\)

\(=-a^4b^4\left(16b^2+24ab+9a^2\right)\)

\(=-a^4b^4\cdot\left(4b+3a\right)^2\)

b) Ta có: \(x^3-6x^2y+12xy^2-8y^3\)

\(=x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\)

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

c) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\)

\(=x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)

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