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

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

\(a^6-a^4+2a^3+2a^2\)

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

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

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

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

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

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

a)  4(2x-3)^2-9(4x^2-9)^2

=[2(2x-3)]^2-[3(4x^2-9)]^2

=(4x-6)^2-(12x^2-27)^2

=(4x-6+12x^2-27)(4x-6-12x^2+27)

=(12x^2+4x-33)(4x-12x^2+21)

b)  a^6-a^4+2a^3+2a^2

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

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

=(a+1)[(a^4)(a-1)+2a^2]

=(a+1)(a^5+a^4+2a^2)

a^6 - a^4 +2a^3 + 2a^2

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

=a^4 (a +1).(a -1) + 2a^2 (a+1)

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

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

Bài làm của mình nếu chưa đúng hoặc có gi sai sót mong các bạn đóng góp ý kiến để hoàn chỉnh hơn nha. Cảm ơn các bạn!

a) \(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)

\(=\left(4x^2-25\right)^2-\left(6x-15\right)^2\)

\(=\left(4x^2-25-6x+15\right)\left(4x^2-25+6x-15\right)\)

\(=\left(4x^2-6x-10\right)\left(4x^2+6x-40\right)\)

\(=\left(4x^2+4x-10x-10\right)\left(4x^2+16x-10x-40\right)\)

\(=\left[4x\left(x+1\right)-10\left(x+1\right)\right]\left[4x\left(x+4\right)-10\left(x+4\right)\right]\)

\(=\left(4x-10\right)\left(x+1\right)\left(4x-10\right)\left(x+4\right)\)

\(=\left(4x-10\right)^2\left(x+1\right)\left(x+4\right)\)

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

b) \(a^6-a^4+2a^3+2a^2\)

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

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

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

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

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