\(x\sqrt{x}-1=\sqrt{x^2}.x-1=\sqrt{x}^3-1=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)

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

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

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

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

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

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

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

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

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

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

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

\(=x^2+2x\cdot\frac{1}{2}+\frac{1}{4}-\left(\frac{\sqrt{23}}{2}i\right)^2\)

\(=\left(x+\frac{1}{2}\right)^2\)\(-\left(\frac{\sqrt{23}}{2}i\right)^2\)

\(\left(x+\frac{1}{2}-\frac{\sqrt{23}}{2}i\right)\left(x+\frac{1}{2}+\frac{\sqrt[]{23}}{2}i\right)\)

\(x^4\ge0;x^2\ge0;4>0\Rightarrow x^4+x^2+4>0\)

đề lỗi rồi

ghi lộn chỉ có x^4+x^2+4 thôi

\(x^3+27x+\left(x+3\right)\left(x-9\right)\)
⇒\(x^3+27x+x^2-6x-27\)
⇒\(x^3+x^2+21x-27\)
Chịu

Sửa đề: \(x^3+27+\left(x+3\right)\left(x-9\right)\)

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

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

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

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

=x¹¹-x²+x²+x+1

=x²(x⁹-1)+(x²+x+1)

=x²[(x³)³-1³)+(x²+x+1)

=x²(x³-1)(x⁶+x³+1)+(x²+x+1)

=x²(x⁶+x³+1)(x-1)(x²+x+1)+(x²+x+1)

Đặt nhân tử chung là x²+x+1 rồi phá hết ngoặc là xong

cá này là bình phương thếu.k thể phân tích thành nhân tử dc nữa

(x+5)²

\(x^2+10x+25=\left(x+5\right)^2\)

(x+5)²