= (x⁴ + x³ + x²) + (1997x² + 1997x + 1997) - (x³ + x² + x)

= x²(x² + x + 1) + 1997(x² + x + 1) - x(x² + x + 1)

= (x² + x + 1)(x² - x + 1997)

bn chép lại đề nha

\(=x^4-x+1997\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+1997\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)

xong nha. chúc bn hc tốt

\(\sqrt{x}\ge0\Rightarrow A=\sqrt{x}-1\ge-1\)

\(\sqrt{x-1}\ge0\Rightarrow B=\sqrt{x-1}-2\ge-2\)

Không tồn tại GTNN của C

=(x⁴+x²+1)+(1996x²+1996x+1996)

=(x²+x+1)(x²-x+1)+1996(x²+x+1)

=(x²+x+1)(x²-x+1+1996)

\(x^4+1997x^{ 2}+1996x+1997\)

\(=\left(x^4-x\right)+\left(1997x^2+1997x+1997\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+1997\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+1997\right]\)

\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)

Ta có :

\(x^4+1997x^2+1996x+1997\)

\(=x^4+1997x^2+1997x-x+1997\)

\(=\left(x^4-x\right)+\left(1997x^2+1997x+1997\right)\)

\(=x\left(x^3-1\right)+1997\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+1997\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)

a) x⁴ + 1997x² + 1996x +1997

= x⁴ + 1997x² + 1997x - x +1997

=(x⁴-x) + (1997x² +1997x+1997)

=x(x³-1) + 1997(x²+x+1)

=x(x-1)(x²+x+1) + 1997(x²+x+1)

=(x²+x+1)(x²-x) + 1997(x²+x+1)

=(x²+x+1)(x²-x+1997)

b) x² -x -2001.2002

=x² - x -2002² +2002

=(x²-2002²)-(x-2002)

=(x-2002)(x+2002) - (x-2002)

=(x-2002)(x+2002+1)

=(x-2002)(x+2003)

c)x⁸ + 98x⁴ +1

= (x⁸+2x⁴+1) + 96x⁴

= (x⁴+1)² + 96x⁴

=[(x⁴+1)² + 2.(x⁴+1).8 + 64x⁴ ]+[32x⁴ - 16x²(x⁴+1)]

=(x⁴+1+8x²)-16x²(-2x²+x⁴+1)

=(x⁴+8x²+1)²- 16x²(x²-1)²

=(x⁴ + 8x² +1)²- [4x(x²-1)]²

=(x⁴+8x²+1)² - (4x³-4x)²

=(x⁴-4x³+8x²+4x+1)(x⁴+4x³+8x²-4x+1)

Pt vô nghiệm

=> dùng hệ số bất định hay phân tích có nhân tử là (x²+x+1)