x² - 4y² + x + 2y

= ( x² - 4y² ) + ( x + 2y )

= ( x - 2y ) ( x + 2y ) + ( x + 2y )

= ( x + 2y ) ( x - 2y + 1 )

\(x^5+x+1\)

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

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

\(x^2\left(x-3\right)-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

=x²(x-3)-4x+3.4

=x²(x-3)-4(x+3)

=x²(x-3)+4(x-3)

=(x-3)(x²+4)

=(x-3)(x²+2²)

=(x-3)(x-2)(x+2)

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

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

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

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

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

= (x^3 - x) + (x^2y - y)

= x(x^2 - 1) + y(x^2 - 1)

= ( x^2 -1)(x+y)

Câu 1:

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

\(=\left(x-y+4\right)\cdot\left(x+y-4\right)\)

Bài 2: 

Sửa đề:  \(x^3-3x^2-10x=0\)

\(\Leftrightarrow x\left(x^2-3x-10\right)=0\)

\(\Leftrightarrow x\left(x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-2\end{matrix}\right.\)

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

=(x-y)-(căn x+căn y)

=(căn x-căn y)(căn x+căn y)-(căn x+căn y)

=(căn x+căn y)(căn x-căn y-1)

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

Dễ thấy \(x^2+1>0\); \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) nên ta không thể phân tích thêm được nữa.

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

\(x^2+5x-36=\left(x-4\right)\left(x+9\right)\)