by Phân tích các đa thức sau thành nhân tử:
\( \eqalign{& a)\,\,2x – 7\sqrt x – 9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,d)\,\,3x + 2\sqrt x – 5 \cr & b)\,\,x – 9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,e)\,\,\,4\sqrt x – x – 4 \cr & c)\,\,x\sqrt x – 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f)\,\,x + \sqrt x – 6 \cr} \)
Lời giải chi tiết:
\(\eqalign{& a)\,\,2x – 7\sqrt x – 9 = 2x + 2\sqrt x – 9\sqrt x – 9 \cr & = 2\sqrt x \left( {\sqrt x + 1} \right) – 9\left( {\sqrt x + 1} \right) \cr & = \left( {\sqrt x + 1} \right)\left( {2\sqrt x – 9} \right). \cr & b)\,x – 9 = {\left( {\sqrt x } \right)^2} – {3^2} \cr & = \left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right). \cr & c)\,\,x\sqrt x – 1 = {\left( {\sqrt x } \right)^3} – 1 \cr & = \left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right). \cr & \cr} \)
\( \eqalign{& d)\,3x + 2\sqrt x – 5 = 3x – 3\sqrt x + 5\sqrt x – 5 \cr & = 3\sqrt x \left( {\sqrt x – 1} \right) + 5\left( {\sqrt x – 1} \right) \cr & = \left( {\sqrt x – 1} \right)\left( {3\sqrt x + 5} \right). \cr & e)\,4\sqrt x – x – 4 = – \left( {x – 4\sqrt x + 4} \right) \cr & = – {\left( {\sqrt x – 2} \right)^2}. \cr & f)\,x + \sqrt x – 6 = x + 3\sqrt x – 2\sqrt x – 6 \cr & = \sqrt x \left( {\sqrt x + 3} \right) – 2\left( {\sqrt x + 3} \right) \cr & = \left( {\sqrt x + 3} \right)\left( {\sqrt x – 2} \right). \cr} \)
