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

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

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

Bài 1:

\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)

\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)

Bài 2:

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

Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9

Bài 4:

 \(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)  

= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)

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

\(\text{nên phép chia:}x^3+x^2-2x\text{ cho:}x+2\text{ ko dư và có thương là:}x^2+x\)

Bài 2:

a: =x(x^2-25)

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

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

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

c: =(2x-3)(4x^2+6x+9)+2x(2x-3)

=(2x-3)(4x^2+8x+9)

bài 1 đâu

A = 6x⁴ - 5x³ + 4x² + 2x - 1

   = 6x⁴ + 3x³ - 8x³ - 4x² + 8x² + 4x - 2x - 1

   = 3x³. ( 2x + 1 ) - 4x² ( 2x + 1 ) + 4x ( 2x + 1 ) - ( 2x + 1 )

   = ( 2x + 1 ) ( 3x³ - 4x² + 4x - 1 )

    = ( 2x + 1 ) ( 3x³ - x² - 3x² + x + 3x - 1 )

     = ( 2x + 1 ) [ x² ( 3x - 1 ) - x ( 3x - 1 ) + ( 3x - 1 ) ]

     = ( 2x + 1 ) ( 3x - 1 ) ( x² - x + 1 )

B = 4x⁴ + 4x³ + 5x² + 8x - 6

    = 4x⁴ - 2x³ + 6x³ - 3x² + 8x² - 4x + 12x - 6

     = 2x³ ( 2x - 1 ) + 3x² ( 2x - 1 ) + 4x ( 2x - 1 ) + 6 ( 2x - 1 )

     = ( 2x - 1 ) ( 2x³ + 3x² + 4x + 6 )

     = ( 2x - 1 ) [ x² ( 2x + 3 ) + 2 ( 2x + 3 ) ]

      = ( 2x - 1 ) ( 2x + 3 ) ( x² + 2 )

C = x⁴ + x³ - 5x² + x - 6

   = x⁴ - 2x³ + 3x³ - 6x² + x² - 2x + 3x - 6 

   = x³ ( x - 2 ) + 3x² ( x - 2 ) + x ( x - 2 ) + 3 ( x - 2 )

   = ( x - 2 ) ( x³ + 3x² + x + 3 )

    = ( x - 2 ) [ x² ( x + 3 ) + ( x + 3 ) ]

    = ( x - 2 ) ( x + 3 ) ( x² + 1 )