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

b) \(10x^4y^2-10x^3y^2-10x^2y^2+10xy^2=10xy^2\left(x^3-x^2-x+1\right)=10xy^2\left(x-1\right)^2\left(x+1\right)\)

a: \(x^4-4x^2-4x-1\)

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

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

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

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

b: \(10x^4y^2-10x^3y^2-10x^2y^2+10xy^2\)

\(=10xy^2\left(x^3-x^2-x+1\right)\)

\(=10xy^2\cdot\left[\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\right]\)

\(=10xy^2\cdot\left(x+1\right)\left(x-1\right)^2\)

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

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

a) x² - 9

= x² - 3²

= (x - 3)(x + 3)

b) 4x² - 1

= (2x)² - 1²

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

c) x⁴ - 16

= (x²)² - 4²

= (x² - 4)(x² + 4)

= (x² - 2²)(x² + 4)

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

d) x² - 4x + 4

= x² - 2.x.2 + 2²

= (x - 2)²

e) x³ - 8

= x³ - 2³

= (x - 2)(x² + 2x + 4)

f) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³

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

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

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

a) Ta có: \(x^4+2x^3-4x-4\)

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

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

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

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

d) Ta có: \(a^3+b^3+c^3-3abc\)

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

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

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

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

a: \(50x^5-8x^3\)

\(=2x^3\left(25x^2-4\right)\)

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

b: \(x^4-5x^2-4y^2+10y\)

\(=\left(x^2-2y\right)\left(x^2+2y\right)-5\left(x^2-2y\right)\)

\(=\left(x^2-2y\right)\left(x^2+2y-5\right)\)

c: \(36a^2+12a+1-b^2\)

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

\(=\left(6a+1-b\right)\left(6a+1+b\right)\)

d: \(x^3+y^3-xy^2-x^2y\)

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

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

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

e: Ta có: \(4x^2+4x-3\)

\(=4x^2+6x-2x-3\)

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

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

f: Ta có: \(9x^4+16x^2-4\)

\(=9x^4+18x^2-2x^2-4\)

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

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

g: Ta có: \(-6x^2+5xy+4y^2\)

\(=-6x^2+8xy-3xy+4y^2\)

\(=-2x\left(3x-4y\right)-y\left(3x-4y\right)\)

\(=\left(3x-4y\right)\left(-2x-y\right)\)

h: Ta có: \(\left(x^2+4x\right)^2+8\left(x^2+4x\right)+15\)

\(=\left(x^2+4x\right)^2+3\left(x^2+4x\right)+5\left(x^2+4x\right)+15\)

\(=\left(x^2+4x+3\right)\cdot\left(x^2+4x+5\right)\)

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

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.

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

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

Ta có:

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

\(=\left[\left(x^4+2x^3\right)-\left(x+2\right)\right]+4\left(x^2+x+1\right)\)

\(=\left[x^3\left(x+2\right)-\left(x-2\right)\right]+4\left(x^2+x+1\right)\)

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

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

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

a: \(x^2-y^2-x-y\)

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

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

f: \(x^3-5x^2-5x+1\)

\(=\left(x+1\right)\left(x^2-x+1\right)-5x\left(x+1\right)\)

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

a: Ta có: \(x^2-4y^2-2x-4y\)

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

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

c: Ta có: \(x^3+2x^2y-x-2y\)

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

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

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

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

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

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

e: Ta có: \(x^3-4x^2-9x+36\)

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

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

f: Ta có: \(x^2-y^2-2x-2y\)

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

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