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

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

vại

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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) \(\frac{1}{64}x^6-125y^3\)

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

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

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

a/ \(E=a^6+a^4+a^2b^2+b^4-b^6\)

\(E=\left[\left(a^2\right)^2+2a^2b^2+\left(b^2\right)^2\right]+\left(a^6-b^6\right)-a^2b^2\)

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

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

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

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

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

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

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

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

       \(a^6+a^4+a^2b^2+b^4-b^6\)

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

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

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

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

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

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

Chúc bạn học tốt.

cảm ơn bạn nhiều nha

a) \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

b) Ta có: \(4x^4+y^4\)

\(=4x^4+y^4+4x^2y^2-4x^2y^2\)

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

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

a, \(x^6-x^4+2x^3+2x^2\)

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

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

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

b, \(4x^4+y^4=\left(2x^2\right)^2+2.2x^2.y^2+\left(y^2\right)^2-4x^2y^2\)

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