by Phân tích đa thức thành nhân tử:
\(\eqalign{& a)\;16{x^4}\left( {x – y} \right) – x + y \cr & c)\;16{x^3} – 54{y^3} \cr & e)\;{x^2} – 9 + \left( {2x + 7} \right)\left( {3 – x} \right) \cr & g)\;4{x^3} – 4{x^2} – x + 1 \cr} \) \(\eqalign{& b)\;2{x^3}y – 2x{y^3} – 4x{y^2} – 2xy \cr & d)\;{x^3} + {x^2} – 4x – 4 \cr & f)\;{x^2} – 2x + 1 – 4{y^2} \cr & h)\;{x^4} – 4{x^3} + 4{x^2} \cr} \)
A. \(\eqalign{& a)\;\left( {2x – 1} \right)\left( {2x + 1} \right)\left( {4{x^2} + 1} \right)\left( {x – y} \right). \cr & b)\;2xy\left( {x – y – 1} \right)\left( {x + y + 1} \right). \cr & c)\;2\left( {2x – 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right). \cr & d)\;\left( {x – 2} \right)\left( {x + 2} \right)\left( {x + 1} \right). \cr & e)\left( {x – 3} \right)\left( { – x – 4} \right). \cr & f)\left( {x – 2y – 1} \right)\left( {x + 2y – 1} \right). \cr & g)\left( {2x – 1} \right)\left( {2x + 1} \right)\left( {x – 1} \right). \cr & h)\;{x^2}{\left( {x – 2} \right)^2}. \cr} \)
B. \(\eqalign{& a)\;\left( {2x + 1} \right)\left( {2x + 1} \right)\left( {4{x^2} + 1} \right)\left( {x – y} \right). \cr & b)\;2xy\left( {x – y – 1} \right)\left( {x + y + 1} \right). \cr & c)\;2\left( {2x + 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right). \cr & d)\;\left( {x – 2} \right)\left( {x + 2} \right)\left( {x + 1} \right). \cr & e)\left( {x – 3} \right)\left( { – x – 4} \right). \cr & f)\left( {x – 2y – 1} \right)\left( {x + 2y – 1} \right). \cr & g)\left( {2x – 1} \right)\left( {2x + 1} \right)\left( {x – 1} \right). \cr & h)\;{x^2}{\left( {x – 2} \right)^2}. \cr} \)
C. \(\eqalign{& a)\;\left( {2x – 1} \right)\left( {2x – 1} \right)\left( {4{x^2} + 1} \right)\left( {x + y} \right). \cr & b)\;2xy\left( {x – y – 1} \right)\left( {x + y + 1} \right). \cr & c)\;2\left( {2x + 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right). \cr & d)\;\left( {x – 2} \right)\left( {x + 2} \right)\left( {x + 1} \right). \cr & e)\left( {x – 3} \right)\left( { – x – 4} \right). \cr & f)\left( {x – 2y – 1} \right)\left( {x + 2y – 1} \right). \cr & g)\left( {2x – 1} \right)\left( {2x + 1} \right)\left( {x – 1} \right). \cr & h)\;{x^2}{\left( {x – 2} \right)^2}. \cr} \)
D. \(\eqalign{& a)\;\left( {2x – 1} \right)\left( {2x + 1} \right)\left( {4{x^2} + 1} \right)\left( {x – y} \right). \cr & b)\;2xy\left( {x – y – 1} \right)\left( {x – y – 1} \right). \cr & c)\;2\left( {2x – 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right). \cr & d)\;\left( {x – 2} \right)\left( {x – 2} \right)\left( {x + 1} \right). \cr & e)\left( {x – 3} \right)\left( { – x – 4} \right). \cr & f)\left( {x – 2y – 1} \right)\left( {x + 2y – 1} \right). \cr & g)\left( {2x – 1} \right)\left( {2x + 1} \right)\left( {x – 1} \right). \cr & h)\;{x^2}{\left( {x – 2} \right)^2}. \cr} \)
Hướng dẫn Chọn đáp án là: A
Phương pháp giải:
– Đặt nhân tử chung, dùng các hằng đẳng thức đáng nhớ hoặc nhóm các hạng tử một cách thích hợp nhằm xuất hiện hằng đẳng thức hoặc nhân tử chung mới.
– Đặt nhân tử chung để được tích các đa thức.
Lời giải chi tiết:
\(\eqalign{& a)\;16{x^4}\left( {x – y} \right) – x + y \cr & = 16{x^4}\left( {x – y} \right) – \left( {x – y} \right) \cr & = \left( {16{x^4} – 1} \right)\left( {x – y} \right) \cr & = \left[ {{{\left( {2x} \right)}^4} – 1} \right]\left( {x – y} \right) \cr & = \left[ {{{\left( {2x} \right)}^2} – 1} \right]\left[ {\left( {2{x^2}} \right) + 1} \right]\left( {x – y} \right) \cr & = \left( {2x – 1} \right)\left( {2x + 1} \right)\left( {4{x^2} + 1} \right)\left( {x – y} \right). \cr} \) \(\eqalign{& b)\;2{x^3}y – 2x{y^3} – 4x{y^2} – 2xy \cr & = 2xy\left( {{x^2} – {y^2} – 2y – 1} \right) \cr & = 2xy\left[ {{x^2} – \left( {{y^2} + 2y + 1} \right)} \right] \cr & = 2xy\left[ {{x^2} – {{\left( {y + 1} \right)}^2}} \right] \cr & = 2xy\left( {x – y – 1} \right)\left( {x + y + 1} \right). \cr} \)
\(\eqalign{& c)\;16{x^3} – 54{y^3} \cr & = 2\left( {8{x^3} – 27{y^3}} \right) \cr & = 2\left[ {{{\left( {2x} \right)}^3} – {{\left( {3y} \right)}^3}} \right] \cr & = 2\left( {2x – 3y} \right)\left[ {{{\left( {2x} \right)}^2} + 2x.3y + {{\left( {3y} \right)}^2}} \right] \cr & = 2\left( {2x – 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right). \cr} \) \(\eqalign{& d)\;{x^3} + {x^2} – 4x – 4 \cr & = \left( {{x^3} + {x^2}} \right) – \left( {4x + 4} \right) \cr & = {x^2}\left( {x + 1} \right) – 4\left( {x + 1} \right) \cr & = \left( {{x^2} – 4} \right)\left( {x + 1} \right) \cr & = \left( {x – 2} \right)\left( {x + 2} \right)\left( {x + 1} \right). \cr} \)
\(\eqalign{& e)\;{x^2} – 9 + \left( {2x + 7} \right)\left( {3 – x} \right) \cr & = \left( {{x^2} – 9} \right) + \left( {2x + 7} \right)\left( {3 – x} \right) \cr & = \left( {x – 3} \right)\left( {x + 3} \right) – \left( {2x + 7} \right)\left( {x – 3} \right) \cr & = \left( {x – 3} \right)\left( {x + 3 – 2x – 7} \right) \cr & = \left( {x – 3} \right)\left( { – x – 4} \right). \cr} \) \(\eqalign{& f)\;{x^2} – 2x + 1 – 4{y^2} \cr & = \left( {{x^2} – 2x + 1} \right) – {\left( {2y} \right)^2} \cr & = {\left( {x – 1} \right)^2} – {\left( {2y} \right)^2} \cr & = \left( {x – 1 – 2y} \right)\left( {x – 1 + 2y} \right) \cr & = \left( {x – 2y – 1} \right)\left( {x + 2y – 1} \right). \cr} \)
\(\eqalign{& g)\;4{x^3} – 4{x^2} – x + 1 \cr & = \left( {4{x^3} – 4{x^2}} \right) – \left( {x – 1} \right) \cr & = 4{x^2}\left( {x – 1} \right) – \left( {x – 1} \right) \cr & = \left( {4{x^2} – 1} \right)\left( {x – 1} \right) \cr & = \left( {{{\left( {2x} \right)}^2} – 1} \right)\left( {x – 1} \right) \cr & = \left( {2x – 1} \right)\left( {2x + 1} \right)\left( {x – 1} \right). \cr} \) \(\eqalign{& h)\;{x^4} – 4{x^3} + 4{x^2} \cr & = {x^2}\left( {{x^2} – 4x + 4} \right) \cr & = {x^2}\left( {{x^2} – 2.2.x + {2^2}} \right) \cr & = {x^2}{\left( {x – 2} \right)^2}. \cr} \)
