a: \(4x^2-x-5=\left(4x-5\right)\left(x+1\right)\)

b: \(x^2-2x-15=\left(x-5\right)\left(x+3\right)\)

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.

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

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

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: \(x^2-y^2-2x-2y\)

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

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

3) \(x^2\left(x+2y\right)-x-2y\)

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

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

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

4) \(x^3-4x^2-9x+36\)

\(=\left(x^3-4x^2\right)-\left(9x-36\right)\)

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

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

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

\(x^2\left(x+2y\right)-x-2y\\ =x^2\left(x+2y\right)-\left(x+2y\right)\\ =\left(x^2-1\right)\left(x+2y\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+2y\right)\\ ---\\ x^3-4x^2-9x+36\\ =x^2\left(x-4\right)-9\left(x-4\right)\\ =\left(x^2-9\right)\left(x-4\right)\\ =\left(x-3\right)\left(x+3\right)\left(x-4\right)\)

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) x3 + 4x2 – 2x – 8

= (x3 + 4x2) - (2x + 8)

= x2(x + 4) - 2(x + 4)

= (x + 4)(x2 - 2)

= (x + 4)(x + √2)(x - √2)

f, \(a^2-5a+14\) không phân tích được thành nhân tử.

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

a: \(=x\left(x^2+4x+4-z^2\right)\)

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

Đề sai rồi bạn