a) \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x+2\right)\left(2-x\right)=1\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(x-2\right)=1\)

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

\(\Leftrightarrow x^3-27-x\left(x^2-2^2\right)=1\)

\(\Leftrightarrow x^3-27-x^3+4x=1\)

\(\Leftrightarrow4x-27=1\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

b) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow\left[\left(x+1\right)-\left(x-1\right)\right]\left[\left(x+1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x-1\right)^2\right]-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow2\left[\left(x+1\right)^2+\left(x^2-1\right)+\left(x-1\right)^2\right]-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow2\left[\left(x^2+2x+1\right)+\left(x^2-1\right)+\left(x^2-2x+1\right)\right]-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow2\left(x^2+2x+1+x^2-1+x^2-2x+1\right)-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow2\left(3x^2+1\right)-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow6x^2+2-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=-\dfrac{6}{12}=-\dfrac{1}{2}\)

\(a,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\\ \Leftrightarrow x^3-3x^2+3x^2-9x+9x-27+\left(x^2+2x\right)\left(2-x\right)-1=0\\ \Leftrightarrow x^3-3x^2+3x^2-9x+9x-27+2x^2-x^3+4x-2x^2-1=0\\ \Leftrightarrow x^3-x^3-3x^2+3x^2+2x^2-2x^2-9x+9x+4x=1+27\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\)

b: 4x^2+4x+1=(2x+1)^2

c: =(6x+1)^2

d: =(3x-4y)^2

e: =(1/2x+2y)^2

f: =-(x^2-10x+25)

=-(x-5)^2

g: =-(16a^4b^6+24a^5b^5+9a^6b^4)

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

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

h: =(5x)^2-2*5x*2y+(2y)^2

=(5x-2y)^2

i: =(5x^2)^2-2*5x^2*y+y^2

=(5x^2-y)^2

\(a,3\left(2a-1\right)+5\left(3-a\right)\)

\(=6a-3+15-5a\)

\(=a-12\)

Thay \(a=\dfrac{-3}{2}\) vào biểu thức trên 

\(a-12\)

\(=\dfrac{-3}{2}-12\)

\(=\dfrac{-27}{2}\)

\(b,25x-4\left(3x-1\right)+7\left(5-2x\right)\)

\(=25x-12x+4+35-14x\)

\(=-1x+39\)

Thay \(x=2,1\) vào biểu thức trên

\(-1x+39\)

\(=-1.2,1+39\)

\(=-2,1+39\)

\(=36,9\)

\(c,4a-2\left(10a-1\right)+8a-2\)

\(=4a-20a+2+8a-2\)

\(=-8a\)

Thay \(a=-0,2\) vào biểu thức trên

\(-8a\)

\(=-8.\left(-0,2\right)\)

\(=1,6\)

\(d,12\left(2-3b\right)+35b-9\left(b+1\right)\)

\(=24-36b+35b-9b-9\)

\(=-10b-15\)

Thay \(b=\dfrac{1}{2}\) vào biểu thức trên 

\(-10b-15\)

\(=-10.\dfrac{1}{2}-15\)

\(=-20\)

a: =6y^3-3y^2-y^2+2y-y+y^2-y^3

=5y^3-3y^2+y

b: =2x^2a-a-2x^2a-a-x^2-ax

=-x^2-ax-2a

c: =2p^3-p^3+1+2p^3+6p^2-3p^5

=3p^3+6p^2-3p^5+1

d: =-3a^3+5a^2+4a^3-4a^2=a^3+a^2

a) \(x^3+6x^2+12x+8\)

\(=x^3+3\cdot x^2\cdot2+3\cdot2^2\cdot x+2^3\)

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

b) \(x^3-3x^2+3x-1\)

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

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

c) \(1-9x+27x^2-27x^3\)

\(=1^3-3\cdot3x\cdot1^2+3\cdot\left(3x\right)^2\cdot1-\left(3x\right)^3\)

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

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

\(=x^3\cdot3\cdot\dfrac{1}{2}\cdot x^2+3\cdot\left(\dfrac{1}{2}\right)^2\cdot x+\left(\dfrac{1}{2}\right)^3\)

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

e) \(27x^3-54x^2y+36xy^2-8y^3\)

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

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

d.

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

$=\frac{x^3}{8}-\frac{3x^2y}{4}+\frac{3xy^2}{2}-y^3$

e.

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

$=\frac{x^3}{8}+\frac{x^2y}{4}+\frac{xy^2}{6}+\frac{y^3}{27}$

f.

$(\frac{2x}{3}-2y)^3=(\frac{2x}{3})^3-3(\frac{2x}{3})^2.2y+3.\frac{2x}{3}(2y)^2-(2y)^3$

$=\frac{8x^3}{27}-\frac{8x^2y}{3}+8xy^2-8y^3$

g.

$(x+y)^3+(x-y)^3=(x^3+3x^2y+3xy^2+y^3)+(x^3-3x^2y+3xy^2-y^3)$

$=2x^3+6xy^2$

Lời giải:

a.
$(3-y)^3=3^3-3.3^2y+3.3y^2-Y63=27-27y+9y^2-y^3$
b.

$(3x+2y^2)^3=(3x)^3+3.(3x)^2(2y^2)+3.3x(2y^2)^2+(2y^2)^3$
$=8y^6+24xy^4+24x^2y^2+8x^3$

c.

$(x-3y^2)^3=x^3-3x^2(3y^2)+3x(3y^2)^2-(3y^2)^3$
$=x^3-9x^2y^2+27xy^4-27y^6$

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

\(b,8-12x+6x^2-x^3=2^3-3.2^2x+3.2x^2-x^3=\left(2-x\right)^3\)

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

\(d,8x^3+12x^2+6x+1\\ =\left(2x\right)^3+3.1\left(2x\right)^2+3.2x.1^2+1^3=\left(2x+1\right)^3\)

a: =4x+2x^2-x^3-2x^2+x^3-4x+3=3

b: =24-4x+2x^2+3x^3-5x^2+4x+3x^2-3x^3=24

c: =x^4-x^3-3x^2+2x-x^4-x^3-3x^2+2x^2+2x+6+4x^2-4x-8

=-2

d: =x^2n-2x^n+x^n-2-x^2n+x^n+2009=2009

\(a,3x\left(5x^2-2x-1\right)=15x^3-6x^2-3x\)

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

\(c,\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)=x^5y-\dfrac{2}{25}x^3y^3-\dfrac{1}{2}x^2y\)

\(d,\dfrac{2}{7}x\left(1,4x-3,5y\right)=\dfrac{2}{5}x^2-xy\)

\(e,\dfrac{1}{2}xy\left(\dfrac{2}{3}x^2-\dfrac{3}{4}x+\dfrac{4}{5}y^2\right)=\dfrac{3}{4}x^3y-\dfrac{3}{8}x^2y+\dfrac{2}{5}xy^3\)

\(f,\left(1+2x-x^2\right)5x=5x+10x^2-5x^3\)