Ta có hằng đẳng thức: 

\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Ta thấy \(\left(x-1\right)+\left(x-2\right)+\left(3-2x\right)=0\)

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

suy ra \(\left(x-1\right)\left(x-2\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x_1=1\\x_2=2\\x_3=\frac{3}{2}\end{cases}}\)

\(S=\frac{29}{4}\).

\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)

a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)

2)
a)
\(=x^3+125\)
\(\)b)\(=1-x^3\)
c)
=\(y^3+27t^3\)
 

a) 4.9.16=2.2.3.3.4.4=2.2.3.3.2.2.2.2=2⁶.3²

b)4.6.9.15.25=2.2.3.2.3.3.3.5.5.5=2³.3⁴.5³=10³.3⁴

a) 2² . 3² . 4²

b) 2² . 6 . 3² . 15 . 5²

cảm ơn dù ko bt đúng sai

a, 4.9.16 = 2².3².4² = 24²

b, 4.6.9.15.25 = 2².2.3.3².3.5.5² = 2³.3⁴.5³ = 3.30³