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

b) \(\left(x^2+x\right)^2+4x^2+4x-12=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16=\left(x^2+x+2\right)^2-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)+1-25=\left(x^2+7x+11\right)^2-5^2=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

a. \(x^2\left(x^2+4\right)-x^2-4\)

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

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

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

b. \(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=x^4+2x^3+5x^2+4x-12\)

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

c. \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\) (*)

Đặt \(t=x^2+7x+10\), ta được

(*) \(=t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=\left(t-4\right)\left(t+6\right)\)

hay \(\left(x^2+7x+6\right)\left(x^2+7x+18\right)\)

a: Ta có: \(x^2\left(x^2+4\right)-x^2-4\)

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

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

b: Ta có: \(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)

\(=\left(x^2+x\right)^2+6\left(x^2+x\right)-2\left(x^2+x\right)-12\)

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

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

c: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)

\(=x^3+27-x^3-54\)

=-27

2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=8x^3+y^3-8x^3+y^3\)

\(=2y^3\)​

\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)

a: Ta có: \(4x\left(x-7\right)-4x^2=56\)

\(\Leftrightarrow4x^2-7x-4x^2=56\)

hay x=-8

b: Ta có: \(12x\left(3x-2\right)-\left(4-6x\right)=0\)

\(\Leftrightarrow36x^2-24x-4+6x=0\)

\(\Leftrightarrow36x^2-18x-4=0\)

\(\text{Δ}=\left(-18\right)^2-4\cdot36\cdot\left(-4\right)=900\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{18-30}{72}=\dfrac{-1}{6}\\x_2=\dfrac{18+30}{72}=\dfrac{2}{3}\end{matrix}\right.\)

c: Ta có: \(4\left(x-5\right)-\left(x-5\right)^2=0\)

\(\Leftrightarrow\left(x-5\right)\left(4-x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)

\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)

b: \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)

c: \(2x-1-x^2\)

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

d: \(x^2+x+\dfrac{1}{4}=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

e: \(9-x^2=3^2-x^2=\left(3-x\right)\left(3+x\right)\)

g: \(\left(x+5\right)^2-4x^2=\left(x+5+2x\right)\left(x+5-2x\right)\)

\(=\left(5-x\right)\left(5+3x\right)\)

h: \(\left(x+1\right)^2-\left(2x-1\right)^2\)

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

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

i: \(=x^2y^2-4xy+4-3\)

\(=\left(xy-2\right)^2-3=\left(xy-2-\sqrt{3}\right)\left(xy-2+\sqrt{3}\right)\)

k: \(=y^2-\left(x-1\right)^2\)

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

l: \(=x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=\left(x+2\right)^3\)

m: \(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2-y^3=\left(2x-y\right)^3\)

`A=x^2-4x+1`
`=x^2-4x+4-3`
`=(x-2)^2-3>=-3`
Dấu "=" xảy ra khi x=2
`B=4x^2+4x+11`
`=4x^2+4x+1+10`
`=(2x+1)^2+10>=10`
Dấu "=" xảy ra khi `x=-1/2`
`C=(x-1)(x+3)(x+2)(x+6)`
`=[(x-1)(x+6)][(x+3)(x+2)]`
`=(x^2+5x-6)(x^2+5x+6)`
`=(x^2+5x)^2-36>=-36`
Dấu "=" xảy ra khi `x=0\or\x=-5`
`D=5-8x-x^2`
`=21-16-8x-x^2`
`=21-(x^2+8x+16)`
`=21-(x+4)^2<=21`
Dấu "=" xảy ra khi `x=-4`
`E=4x-x^2+1`
`=5-4+4-x^2`
`=5-(x^2-4x+4)`
`=5-(x-2)^2<=5`
Dấu "=" xảy ra khi `x=5`

A= x² - 4x +1

   = x² - 4x + 4 - 3

   = (x-2)² -3

Ta có (x-2)² ≥ 0 ∀ x

    ⇒ (x-2)² -3 ≥ -3 ∀ x

Vậy A_Min= -3 tại x=2

B= 4x²+4x+11

  = 4x²+4x+1+10

  = (2x+1)²+10

Ta có (2x+1)² ≥ 0 ∀ x

     ⇒ (2x+1)²+10 ≥ 10 ∀ x

Vậy B_Min=10 tại x= \(\dfrac{-1}{2}\)

C=(x-1)(x+3)(x+2)(x+6)

  = (x-1)(x+6)(x+3)(x+2)

  = (x²+5x-6) (x²+5x+6)

  = (x²+5x)² -36

Ta có (x²+5x)² ≥ 0 ∀ x
  ⇒ (x²+5x)² -36 ≥ -36 ∀ x

Vậy C_Min=-36 tại x=0 hoặc x= -5

\(a,=\left(3x-5\right)\left(3x+3\right)=3\left(x+1\right)\left(3x-5\right)\\ b,=\left(5x-4-7x\right)\left(5x-4+7x\right)=\left(-2x-4\right)\left(12x-4\right)\\ =-8\left(x+2\right)\left(x-3\right)\\ c,=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\\ =\left(x+14\right)\left(3x-4\right)\\ d,=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\\ =\left(x+5\right)\left(5x-3\right)\\ e,=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\\ =\left(4x+7\right)\left(8x+11\right)\\ f,=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\\ =\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\\ =\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\\ g,=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\\ =\left(a-b\right)\left(x-y\right)\left(a+b\right)\left(x+y\right)\)

\(h,=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\\ =\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\\ =\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)

a: \(\left(3x-1\right)^2-16\)

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

\(=\left(3x+3\right)\left(3x-5\right)\)

\(=3\left(x+1\right)\left(3x-5\right)\)

b: \(\left(5x-4\right)^2-49x^2\)

\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)

\(=\left(-2x-4\right)\left(12x-4\right)\)

\(=-8\left(x+2\right)\left(3x-1\right)\)