a: \(x^4+x^2+2x+6\)

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

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

x⁸+3x⁴+4

= x⁸+4x⁴+4-x⁴

= (x⁴+2)²-(x²)²

= (x⁴+2-x²).(x⁴+2+x²)

\(x^8+3x^4+4=(x^8+3x^4)+4=x^4(x^2+3)+4=(x^2+3)(x^4+4)\)

câu hỏi hay......nhưng tui xin nhường cho các bn khác

Hãy tích đúng cho tui nha

THANKS

\(x^8+3x^4+4\)

\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)

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

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

\(4x^4+4x^3+5x^2+2x+1\)

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

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

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

Câu b đề sai nha bạn.

a) \(x^3+9x^2+27x+27=\left(x+3\right)^3\)

b) \(3\sqrt{3x^3}+18x^2+12\sqrt{3x}+8=\left(\sqrt{3x}+2\right)^3\)

c) \(\dfrac{1}{4}-x^2=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)

=4x²-4=4(x²-1)=4(x-1)(x+1)

4(x²-1)

1: \(-x^2+2x+8\)

\(=-\left(x^2-2x-8\right)\)

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

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