\(=4x^2-28x-8x+56=4x\left(x-7\right)-8\left(x-7\right)=4\left(x-2\right)\left(x-7\right)\)

Lời giải:
\(x^3(x^2-7)^2-36x=x[x^2(x^2-7)^2-36]\\ =x[(x^3-7x)^2-6^2]=x(x^3-7x-6)(x^3-7x+6)\\ =x[x^2(x-3)+3x(x-3)+2(x-3)][x^2(x-2)+2x(x-2)-3(x-2)]\\ =x(x-3)(x^2+3x+2)(x-2)(x^2+2x-3)\\ =x(x-3)(x+1)(x+2)(x-2)(x-1)(x+3)\)

2) \(x^4-5x^2+4\)

\(=x^4-x^2-4x^2+4\)

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

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

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

x^3.(x^2-7)^2-36x

=x(x^6-14x^4+49x^2-36)

=x.[x^4(x^2-1)-13x^2(x^2-1)+36(x^2-1)

=x(x-1)(x+1)(x^4-13X^2+36)

=x(x-1)(x+1)[x^2(x^2-4)-9(x^2-4)]

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

Ta có : x³ . ( x² - 7 )² - 36x

=> x ( x⁶ - 14x⁴ + 49x² - 36 )

=> x [ x⁴ ( x² - 1 ) - 13x² ( x² - 1 ) + 36 ( x² - 1 )

=> x ( x - 1 ) ( x + 1 ) ( x⁴ - 13x² + 36 )

=> x ( x - 1 ) ( x + 1 ) [ x² ( x² - 4 ) - 9 ( x² - 4 ) ]

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

(x²+9)²-36x²

=(x²+9)²-(6x)²

=(x²+9+6x).(x²+9-6x)

\(x^3\left(x^2-7\right)^2-36x=x^3\left(x^4-14x^2+49\right)-36x\)

=\(x^7-14x^5+49x^3-36x\)

=\(x^7-x^6+x^6-x^5-13x^5+13x^4-13x^4+13x^3+36x^3-36x\)

=\(x^6\left(x-1\right)+x^5\left(x-1\right)-13x^4\left(x-1\right)-13x^3\left(x-1\right)+36x\left(x^2-1\right)\)

=\(x\left(x-1\right)\left(x^5+x^4-13x^3-13x^2+36x+36\right)\)

=\(x\left(x-1\right)\left[x^4\left(x+1\right)-13x^2\left(x+1\right)+36\left(x+1\right)\right]\)

=\(x\left(x-1\right)\left(x+1\right)\left(x^4-13x^2+36\right)\)

đặt x^2 =a (a>=0) thì xét đa thức \(x^4-13x^2+36=a^2-13a+36\)

xét \(\Delta=b^2-4ac=169-4.36=25\)

\(\Delta>0\)→phương trình có 2 nghiệm riêng biệt là \(\left[\begin{array}{nghiempt}a_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{13+5}{2}=9\\a_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{13-5}{2}=4\end{array}\right.\)(t/m a>=0)

vậy bt ban đầu :\(x\left(x-1\right)\left(x+1\right)\left(x^2-4\right)\left(x^2-9\right)\)

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

\(3x^3+3x^2-36x\)

\(=3x\left(x^2+x-12\right)\)

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

4x² - 36x + 56 

= 4(x² - 9x + 14)

= 4(x² - 2x - 7x + 14)

= 4[x(x - 2) - 7(x - 2)]

= 4(x - 2)(x - 7)

\(4x^2\)−36x+56=04x2−36x+56

⇒4(x2−9x+14)=0⇒4(x2−9x+14)

⇒4(x2−7x−2x+14)=0⇒4(x2−7x−2x+14)

⇒4x(x−2)−7(x−2)=0⇒4x(x−2)−7(x−2)

⇒4(x−7)(x−2)=0⇒4(x−7)(x−2)

⇒(x−7)(x−2)=0⇒(x−7)(x−2)

⇒[x−7=0x−2=0⇒[x−7=0x−2=0

⇒x=7;x=2⇒x=7;x=2.