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

=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

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

\(x^8+x+1\)

\(=\left(x^8-x^5\right)+\left(x^5-x^2\right)+\left(x^2+x+1\right)\)

\(x^5\left(x^3-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)

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

\(=\left(x^6-x^5\right)\left(x^2+x+1\right)+\left(x^3-x^2\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

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

chắc là x^3 + 27x phải là x^3 + 27
cô tôi nhầm đề rồi

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

`#3107.101107`

`x(y - 1) + 3(y - 1)`

`= (x + 3)(y - 1)`

x(y-1)+3(y-1)

=(y-1)(x+3)

Giải thích: đặt y-1 ra làm chung .... đa thức còn x+3