Câu 1:

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

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

a: \(x^3-9x^2+6x+16\)

\(=x^3-8x^2-x^2+8x-2x+16\)

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

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

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

b: \(x^3-x^2-x-2\)

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

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

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

c: \(x^3+x^2-x+2\)

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

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

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

d: \(x^3-6x^2-x+30\)

\(=x^3+2x^2-8x^2-16x+15x+30\)

\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-8x+15\right)\)

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

e: Sửa đề: \(x^3-7x-6\)

\(=x^3-x-6x-6\)

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

\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)

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

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

f: \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)

\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)

g: \(2x^3-x^2+5x+3\)

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

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

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

h: \(\left(x^2-3\right)^2+16\)

\(=x^4-6x^2+9+16\)

\(=x^4-6x^2+25\)

\(=x^4+10x^2+25-16x^2\)

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

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

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

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

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

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

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

=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)

\(=x^2-6x+9-2=\left(x-3\right)^2-2=\left(x-3-\sqrt{2}\right)\left(x-3+\sqrt{2}\right)\)

\(=\left(x^2-6x+9\right)-2=\left(x-3\right)^2-\sqrt{2^2}=\left(x-3-\sqrt{2}\right)\left(x-3+\sqrt{2}\right)\)

=x2−6x+9−2=(x−3)2−2=(x−3−√2)(x−3+√2)

bạn có thể viết rõ hơn ko

x^4+x^2y^2+y^4

=x^4+2x^2y^2+y^4-x^2y^2

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

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

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

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

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

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

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

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

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

Bài 2: 

Sửa đề:  \(x^3-3x^2-10x=0\)

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

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

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