\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)

a

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

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) \(-x^3+6x^2y-12xy^2+8y^3\)

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

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

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

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

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

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

\(8x^2y-18y\)

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

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

\(3x^3+6x^2+3x-12xy^2\)

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

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

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

3x³ + 6x² + 3x - 12xy² 

= 3x(x² + 2x + 1 - 4y²) 

= 3x[(x + 1)² - (2y)²]

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

\(3x^3+6x^2+3x-12xy^2\)

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

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

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

\(a.=x^3+3x^2y+3x^2y+9xy^2+3xy^2+9y^3\)

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

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

\(b.=9x^3+3x^2y+9x^2y+3xy^2+3xy^2+y^3\)

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

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

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

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