Đề có gì đó sai sai

1: \(a^2-4b^2-2a-4b\)

\(=\left(a-2b\right)\left(a+2b\right)-2\left(a+2b\right)\)

\(=\left(a+2b\right)\left(a-2b-2\right)\)

2: \(x^3+2x^2-2x-1\)

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

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

1: 2a+2b=2(a+b)

2: 2a+4b+6c

=2*a+2*2b+2*3c

=2(a+2b+3c)

3: \(-7a-14ab-21b=-7\left(a+2ab+3b\right)\)

4: \(2ax-2ay+2a=2a\left(x-y+1\right)\)

5: \(=3a\cdot ax-3a\cdot2ay+3a\cdot4=3a\left(ax-2ay+4\right)\)

6: \(=2\cdot2ax-2\cdot ay-2\cdot1=2\cdot\left(2ax-ay-1\right)\)

7: =a^2-(2b)^2

=(a-2b)(a+2b)

8: =(5a)^2-1^2

=(5a-1)(5a+1)

9: =9(16a^2-9)

=9(4a-3)(4a+3)

\(=4\left[a^2-\left(b^2-4bc+4c^2\right)\right]\)

\(=4\left[a^2-\left(b-2c\right)^2\right]\)

\(=4\left(a-b+2c\right)\left(a+b-2c\right)\)

\(4a^2-4b^2+16bc-16c^2\)

\(=4a^2-\left(4b^2-16bc+16c^2\right)\)

\(=\left(2a\right)^2-\left[\left(2b\right)^2-2.2b.4c+\left(4c\right)^2\right]\)

\(=\left(2a\right)^2-\left(2b-4c\right)^2\)

\(=\left(2a+2b-4c\right)\left(2a-2b+4c\right)\)

\(=4\left(a+b-c\right)\left(a-b+c\right)\)

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

\(a^2-b^2-2x\left(a-b\right)\)

\(=\left(a-b\right)\left(a+b\right)-2x\left(a-b\right)\)

\(=\left(a-b\right)\left(a+b-2x\right)\)

\(-x^2-2x+15=-\left(x^2+2x-15\right)\)

\(=-\left[\left(x^2-3x\right)+\left(5x-15\right)\right]\)

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

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

Ta có: -x^2-2x+15 = -x^2-5x+3x+15 = -x(x+5)+3(x+5) = (3-x)(x+5).
Hok tốt ~~~

-x² - 2x + 15

= -x² + 3x - 5x + 15

= ( 3x - x² ) + ( 15 - 5x )

= x( 3 - x ) + 5( 3 - x )

= ( 3 - x )( x + 5 )

câu a sai đề ko bạn

\(a,a^2-2a-4b^2-4b\)

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

\(=\left(a-2b\right)\left(a+2b\right)-2\left(a+2b\right)\)

\(=\left(a+2b\right)\left(a-2b-2\right)\)

\(b,x^3-2x^2+4x-8\)

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

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

\(c,x^3+36x-12x^2\)

\(=x^3-6x^2-6x^2+36x\)

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

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

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

\(d,5a^2+3\left(a+b\right)^2-5b^2\)

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

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

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

\(=\left(a+b\right)\left[5\left(a-b\right)+3\left(a+b\right)\right]\)

\(=\left(a+b\right)\left(5a-5b+3a+3b\right)\)

\(=\left(a+b\right)\left(8a-2b\right)\)

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

\(e,x^3-3x^2+3x-1-y^3\)

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

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

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

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

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

#Urushi☕

\(c.\\ x^3+36x-12x^2\\ =x\left(x^2-12x+36\right)\\ =x.\left(x^2-2.x.6+6^2\right)\\ =x.\left(x-6\right)^2\\ ---\\ d.\\ 5a^2+3\left(a+b\right)^2-5b^2\\ =\left(5a^2-5b^2\right)+3\left(a+b\right)^2\\ =5.\left(a^2-b^2\right)+3.\left(a+b\right)\left(a+b\right)\\ =5\left(a+b\right)\left(a-b\right)+3\left(a+b\right)\left(a+b\right)\\ =\left(a+b\right)\left(5a-5b+3a+3b\right)\\ =\left(a+b\right)\left(8a-2b\right)\\ =2\left(a+b\right)\left(4a-b\right)\)

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

\(a.\\ a^2-2a-4b^2-4b\\ =\left(a^2-2a+1\right)-\left(4b^2-4b+1\right)\\ =\left(a-1\right)^2-\left(2b-1\right)^2\\ =\left[\left(a-1\right)+\left(2b-1\right)\right].\left[\left(a-1\right)-\left(2b-1\right)\right]\\ =\left(a+2b-2\right)\left(a-2b\right)\)

\(b.\\ x^3-2x^2+4x-8=x^2\left(x-2\right)+4\left(x-2\right)\\ =\left(x^2+4\right)\left(x-2\right)\)

a: =64x^4+16x^2y^2+y^4-16x^2y^2

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

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

b: =x^8+2x^4+1-x^4

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

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

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

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

c: =(x+1)(x^2-x+1)+2x(x+1)

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

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

d: =(x^2-1)(x^2+1)-2x(x^2-1)

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

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

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

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.

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

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