x⁸ + x⁴ + 1

= x⁸ + 2x⁴ + 1 - x⁴

= (x⁴ + 1)² - x⁴

= (x⁴ + 1)² - (x²)²

= (x⁴ + 1 + x²)(x⁴ + 1 - x²)

= (x⁴ + x² + 1)(x⁴ - x² + 1)

a. $6x^2-11x=x(6x-11)$
b. $x^7+x^5+1=(x^7-x)+(x^5-x^2)+x+x^2+1$

$=x(x^6-1)+x^2(x^3-1)+(x^2+x+1)$
$=x(x^3-1)(x^3+1)+x^2(x^3-1)+(x^2+x+1)$
$=(x^3-1)(x^4+x+x^2)+(x^2+x+1)$

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

$=(x^2+x+1)(x^5-x^4+x^3-x+1)$

c.

$x^8+x^4+1=(x^4)^2+2.x^4+1-x^4$

$=(x^4+1)^2-(x^2)^2$

$=(x^4+1-x^2)(x^4+1+x^2)$

$=(x^4+1-x^2)(x^4+2x^2+1-x^2)$

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

$=(x^4-x^2+1)(x^2+1-x)(x^2+1+x)$

d.

$x^3-5x+8-4=x^3-5x+4$

$=x^3-x^2+x^2-x-(4x-4)$

$=x^2(x-1)+x(x-1)-4(x-1)=(x-1)(x^2+x-4)$

e.

$x^5+x^4+1=(x^5-x^2)+(x^4-x)+x^2+x+1$

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

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

$=(x^2+x+1)[(x-1)(x^2+x)+1]$

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

\(a,=\left(5x^3+10x\right)+\left(x^4-4\right)\\ =5x\left(x^2+2\right)+\left(x^2+2\right)\left(x^2-2\right)\\ =\left(x^2+2\right)\left(x^2+5x-2\right)\\ b,=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+2xy+y-xz-yz+z^2-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(c,=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\\ d,=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\\ e,=\left(x^{10}+x^9+x^8\right)-\left(x^9+x^8+x^7\right)+\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^{10}-x^7+x^5-x^4+x^3-x+1\right)\)

a: =x^4+2x^2+5x^3+10x-2x^2-4

=(x^2+2)(x^2+5x-2)

b; =(x+y)^3+z^3-3xy(x+y)-3xyz

=(x+y+z)*(x^2+2xy+y^2-xz-yz+z^2)-3xy(x+y+z)

=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)

c: =x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1

=(x^2+x+1)(x^6-x^5+x^3-x^2+1)

e) Ta có: \(x^4-2x^3+2x-1\)

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

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

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

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

h) Ta có: \(3x^2-3y^2-2\left(x-y\right)^2\)

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

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

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

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

a) Ta có: \(x^2-y^2-2x-2y\)

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

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

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

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

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

c) Ta có: \(x^3-4x^2-9x+36\)

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

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

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

d) Ta có: \(x^4+2x^3+2x-1\)

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

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

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

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

c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)

d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)

a) Ta có: \(x^2-2x-4y^2-4y\)

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

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

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

b) Ta có: \(x^3+2x^2+2x+1\)

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

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

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

d) Ta có: \(a^6-a^4+2a^3+2a^2\)

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

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

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

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

c) Ta có: \(x^3-4x^2+12x-27\)

\(=\left(x^3-27\right)-\left(4x^2-12x\right)\)

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

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

1/(x+2)² -(3x-1)²=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x²+12x

2/(x⁴+x²)(-2x³-2x)=x²(x²+1)-2x(x²+1)=(x²+1)(x²-2x)

Lời giải:

a. Không phân tích được thành nhân tử

b. \(a^4+a^2-22=(a^2+\frac{1}{2})^2-\frac{89}{4}=(a^2+\frac{1-\sqrt{89}}{2})(a^2+\frac{1+\sqrt{89}}{2})\)

(thông thường nhân tử là số hữu tỉ, phân tích kiểu này như cố để thành nhân tử cũng không hợp lý lắm, bạn coi lại đề)

c.

$x^4+4x^2-5=(x^4-x^2)+(5x^2-5)$

$=x^2(x^2-1)+5(x^2-1)=(x^2-1)(x^2+5)=(x-1)(x+1)(x^2+5)$

Nếu sửa như bạn nói thì làm như sau:

a. 

$a^4+a^2+1=(a^2+2a^2+1)-a^2=(a^2+1)^2-a^2=(a^2+1-a)(a^2+1+a)$
b.

$a^4+a^2-2=(a^4-1)+(a^2-1)=(a^2-1)(a^2+1)+(a^2-1)$

$=(a^2-1)(a^2+1+1)=(a-1)(a+1)(a^2+2)$