Bài 43 trang 20 SGK Toán 8 tập 1

 Phân tích các đa thức sau thành nhân tử:

a) \({x^2} + 6x + 9\);

b) \(10x - 25 - {x^2}\);

c) \(8{x^3}-\dfrac{1}{8}\);

d) \(\dfrac{1}{25}{x^2} - 64{y^2}\)

\(\begin{array}{l} a)\;\;{x^2} + 6x + 9 = {x^2} + 2.x.3 + {3^2}\\ = {\left( {x + 3} \right)^2}.\\ b)\;10x - 25 - {x^2} \\= - \left( {{x^2} - 10x + 25} \right) \\=-(x^2-2.x.5+5^2)\\= - {\left( {x - 5} \right)^2}.\\ c)\;8{x^3} - \dfrac{1}{8} = {\left( {2x} \right)^3} - {\left( {\dfrac{1}{2}} \right)^3}\\ = \left( {2x - \dfrac{1}{2}} \right)\left[ {{{\left( {2x} \right)}^2} + 2x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right]\\ = \left( {2x - \dfrac{1}{2}} \right)\left( {4{x^2} + x + \dfrac{1}{4}} \right).\\ d)\;\dfrac{1}{{25}}{x^2} - 64{y^2} = {\left( {\dfrac{1}{5}x} \right)^2} - {\left( {8y} \right)^2}\\ = \left( {\dfrac{1}{5}x - 8y} \right)\left( {\dfrac{1}{5}x + 8y} \right). \end{array}\)