đề sai bét            

x³-x²-4=x³-2x²+x²-4=x²(x-2)+(x-2)(x+2)=(x-2)(x²+x+2)

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

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

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

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

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

\(\left(x-3\right).\left(x+3\right)\)\(+\left(x-3\right)\left(x+4\right)\)=\(\left(x-3\right)\left(x+3+x+4\right)=\left(x-3\right)\left(2x+7\right)\)

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

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

\(\frac{x^2-5}{3x}-\frac{2}{3}=\frac{x^2-2x-5}{3x}=\frac{\left(x-1+\sqrt{6}\right)\left(x-1-\sqrt{6}\right)}{3x}\)

Sửa đề: \(\frac13x^3+\frac72x^2+2x+1\)

\(\frac13x^3+\frac72x^2+2x+1\)

\(=\frac16\cdot21x^3+\frac16\cdot21x^2+\frac16\cdot12x+\frac16\cdot6\)

\(=\frac16\left(2x^3+21x^2+12x+6\right)\)

Biểu thức này không phân tích thành nhân tử.

\(5-7x^2=\left(\sqrt{5}\right)^2-\left(x\sqrt{7}\right)^2\)

\(=\left(\sqrt{5}-x\sqrt{7}\right)\left(\sqrt{5}+x\sqrt{7}\right)\)

\(3+4x=\left(\sqrt{3}\right)^2-\left(2\sqrt{x}\right)^2\) ( do x<0 )

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