\(x^4+6x^3+11x^2+6x+1\)

\(=x^4+3x^3+x^2+3x^3+9x^2+3x+x^2+3x+1\)

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

a) ( x 2  – 4x + 1)( x 2  – 2x + 3).

b) ( x 2  + 5x – 1)( x 2  + x – 1).

c: \(x^4+x^3-4x^2+x+1\)

\(=x^4-x^3+2x^3-2x^2-2x^2+2x-x+1\)

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

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

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

(x + y)³ - 1 - 3xy(x + y - 1)

= x³ + 3x²y + 3xy² + y³ - 1 - 3x²y - 3xy² + 3xy

= x³ - 1 + 3xy

= x(x² + 3y) - 1

k bt lm nx r :v

\(\left(x+y\right)^3-1-3xy\left(x+y-1\right) \)

\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)

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

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

B = (x + 3)(x - 1)(x - 5)(x + 15) + 64x²

B = x⁴ + 12x³ - 58x² - 180x + 225 + 64x²

B = x⁴ + 12x³ + 6x² - 180x + 225

\(x^4-27x=x\left(x^3-27\right)=x\left(x-3\right)\left(x^2+3x+9\right)\)

\(27x^5+x^2=x^2\left(27x^3+1\right)=x^2\left[\left(3x\right)^3+1^3\right]=x^2\left(3x+1\right)\left(9x^2-3x+1\right)\)

a) x⁴-27x=x(x³-27)=x(x-3)(x²-3x+9)

b) 27x⁵+x²=x²(27x³+1)=x²(3x+1)(9x²-3x+1)

Bài 1:

\(a,x^4+5x^2+9\\=(x^4+6x^2+9)-x^2\\=[(x^2)^2+2\cdot x^2\cdot3+3^2]-x^2\\=(x^2+3)^2-x^2\\=(x^2+3-x)(x^2+3+x)\)

\(b,x^4+3x^2+4\\=(x^4+4x^2+4)-x^2\\=[(x^2)^2+2\cdot x^2\cdot2+2^2]-x^2\\=(x^2+2)^2-x^2\\=(x^2+2-x)(x^2+2+x)\)

\(c,2x^4-x^2-1\\=2x^4-2x^2+x^2-1\\=2x^2(x^2-1)+(x^2-1)\\=(x^2-1)(2x^2+1)\\=(x-1)(x+1)(2x^2+1)\)

Bài 2:

\(a,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\cdot\left[\left(x+2\right)\left(x+3\right)\right]=120\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=120\) (1)

Đặt \(x^2+5x+5=y\), khi đó (1) trở thành:

\(\left(y-1\right)\left(y+1\right)=120\)

\(\Leftrightarrow y^2-1=120\)

\(\Leftrightarrow y^2=121\)

\(\Leftrightarrow\left[{}\begin{matrix}y=11\\y=-11\end{matrix}\right.\)

+, TH1: \(y=11\Leftrightarrow x^2+5x+5=11\)

\(\Leftrightarrow x^2+5x-6=0\)

\(\Leftrightarrow x^2-x+6x-6=0\)

\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\left(\text{nhận}\right)\)

+, TH2: \(y=-11\Leftrightarrow x^2+5x+5=-11\)

\(\Leftrightarrow x^2+5x+16=0\)

\(\Leftrightarrow\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{25}{4}+16=0\)

\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)

Ta thấy: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}>0\forall x\)

Mà \(\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)

\(\Rightarrow\) loại

Vậy \(x\in\left\{1;-6\right\}\).

\(b,\) Đề thiếu vế phải rồi bạn.

a: =x^3(x-y)+(x-y)

=(x-y)(x^3+1)

=(x-y)(x+1)(x^2-x+1)

b: =(a-1)^2-9b^2

=(a-1-3b)(a-1+3b)

a) 3xy² - 3x³ - 6xy + 3x 

=3x (y² - x² - 2y +1)

= 3x [ (y-1)² -x² ]

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

b) 3x² +11x+6

= 3 x² +9x +2x +6

=3x (x+3)+2(x+3)

= (x+3)(3x+2)

c) -x³ - 4xy² + 4x²y +16x

= -x (x² + 4y² - 4xy -16 )

= -x [(x -2y)² - 4² ]

= -x(x-2y-4)(x-2y+4)