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

a6, a4 là số mũ hay hệ số vậy bn

Đặt \(A=2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4\)

\(A=-\left(a^4+b^4+c^4-2\left(ab\right)^2-2\left(bc\right)^2-2\left(ca\right)^2\right)\)

\(A=-\left(a^4+b^4+c^4-2\left(ab\right)^2-2\left(bc\right)^2+2\left(ca\right)^2-4\left(ca\right)^2\right)\)

Áp dụng hàng đẳng thức \(\left(a^2-b^2+c^2\right)=a^4+b^4+c^4-2\left(ab\right)^2-2\left(bc\right)^2+2\left(ca\right)^2\):

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

\(A=-\left(a^2-b^2+c^2-2ca\right)\left(a^2-b^2+c^2+2ca\right)\)

2222222222222a+257222222222222222222222222222222222222222222222222222222222222222222222222222222222222222a=?

1)  \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)

2)  \(x^3-9x^2+6x+16\)

\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)

\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)

\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)

3)   \(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)

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

4)  \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)

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

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

5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)

\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)

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

\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)

gửi phần này trước còn lại làm sau !!! tk mk nka !!!

nhiều thế

6) \(\left(x+y\right)^2-\left(x+y\right)-12\)\(=\left(x+y\right)^2-2\cdot\frac{1}{2}\left(x+y\right)+\frac{1}{4}-\frac{49}{4}\)

\(=\left(x+y-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)\(=\left(x+y-\frac{1}{2}-\frac{7}{2}\right)\left(x+y-\frac{1}{2}+\frac{7}{2}\right)\)

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

7)   \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)          (NHÂN x + 2 vs x +  5  và  x + 3 vs x + 4 )

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

ĐẶT   \(x^2+7x+11=y\)   ta được :  

\(\left(y+1\right)\left(y-1\right)-24=y^2-1-24\)

\(=y^2-25=\left(y-5\right)\left(y+5\right)\)

8)  \(4x^4-32x^2+1=4x^4+4x^2+1-36x^2\)

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

9) sai đề rùi bạn ơi ! đề đúng nè 

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

Ta thấy :  

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

Thay vào biểu thức bài cho ta được : 

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

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

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

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

bài ở trên câu 3 : kết luận là  \(\left(x-3\right)\left(x^2-x-6\right)\)bạn sửa lại giúp mk nka !!! Th@nk !!! Tk Mk vs  

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

\(4a^4+b^4\)

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

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

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

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

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.

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

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

\(9a^2b^2-b^4+6b^3-9b^2\\ =b^2\left(9a^2-b^2+6b-9\right)\\ =b^2\left[9a^2-\left(b-3\right)^2\right]\\ =b^2\left(3a-b+3\right)\left(3a+b-3\right)\)