a/ \(\left(a^2-b^2+1\right)\left(a^2-ab+b^2\right)\left(a^2+ab+b^2\right)\)

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

       \(a^6+a^4+a^2b^2+b^4-b^6\)

\(=\left(a^6-b^6\right)+\left(a^4+a^2b^2+b^4\right)\)

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

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

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

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

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

Chúc bạn học tốt.

cảm ơn bạn nhiều nha

a) \(9\left(a+b\right)^2-4\left(a-2b\right)^2\)

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

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

\(=\left(a+7b\right)\left(5a-b\right)\)

b) \(9x^6-12x^7+4x^8\)

\(=x^6\left(9-12x+4x^2\right)\)

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

c) \(8x^6-27y^3\)

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

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

d) \(\frac{1}{64}x^6-125y^3\)

\(=\left(\frac{1}{4}x^2\right)^3-\left(5y\right)^3\)

\(=\left(\frac{1}{4}x^2-5y\right)\left(\frac{1}{16}x^4+\frac{5}{6}xy+25y^2\right)\)

vại

fdvfdverberrgtrgrgg

a,

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

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

= \(\left(\left(a-\sqrt{2b}\right)\left(a+\sqrt{2b}\right)\right)\left(a^2+2b\right)\)

c, 

=\(4x^4+20x^2+25\)

=\(\left(2x^2\right)^2+2.2x^2.5+5^2\)

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

d,

=\(8x^6-27y^3\)

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

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

Câu b đề ghi ko rõ lắm

1.=[(1/2)a^2)^2-2.(1/2)a^2b+b^2
=[(1/2)a^2-b]^2
2.=2a^2+2b^2-2-a^2c+c-b^2c
=2(a^2+b^2-a)-c(a^2+b^2-1)
=(2-c)(a^2+b^2-1)

2a²b²+2b²c²+2a²c²-a⁴-b⁴-c⁴

=4a²c²-(a⁴+b⁴+c⁴-2a²b²+2a²c²-2b²c²)

=4a²c²-(a²-b²+c²)²

=(2ac+a²-b²+c²)(2ac-a²+b²-c²)

=[(a+c)²-b²][b²-(a-c)²]

=(a+b+c)(a+c-b)(b+a-c)(b-a+c)

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

ko bik

a^4+b^4+c^4-2a^2b^2+2b^2c^2-2a^2c^2-4b^2c^2

=(a^2-b^2-c^2)-4b^2c^2

=(a^2-b^2-c^2-2bc)(a^2-b^2-c^2+2bc)

=(a-b-c)(a+b+c)(a-b+c)(a+b-c)

a⁴+b⁴+c⁴- 2a²b²- 2b²c²- 2c²a²

= (a²-b²-c²)²