Phân tích các đa thức sau thành nhân tử:

$$\begin{gathered} a) x^2+6x+9 ; b) 10x-25-x^2; \\ c) 8x^3-\frac{1}{8}; d) \frac{1}{25}{x}^2 - 64y^2; \end{gathered}$$

Giải:

$$\begin{gathered} a) x^2+6x+9 \\ =x^2+2.x.3+3^2 \\ ={(x+3)}^2 \end{gathered}$$

$$\begin{gathered} b) 10x-25-x^2 \\ =-(-10x+25+x^2) \\ =-(25-10x+x^2) \\ =-(5^2-2.5.x-x^2) \\ =-{(5-x)}^2 \end{gathered}$$

$$\begin{gathered} c) 8x^3-\frac{1}{8} \\ ={\left(2x\right)}^3-{\left(\frac{1}{2}\right)}^3 \\ =(2x-\frac{1}{2}\left)\right[{\left(2x\right)}^2+2x.\frac{1}{2}+{\left(\frac{1}{2}\right)}^2] \\ =(2x-\frac{1}{2}\left)\right(4x^2+x+\frac{1}{4}) \end{gathered}$$

$$\begin{gathered} d) \frac{1}{25}{x}^2 - 64y^2 \\ ={\left(\frac{1}{5}x\right)}^2-{\left(8y\right)}^2=(\frac{1}{5}x+8y\left)\right(\frac{1}{5}x-8y) \end{gathered}$$