Đề bài
Rút gọn các biểu thức sau:
a] \[[x + 3][x - 2] + {[x - 3]^2}\] ;
b] \[{[x + 2]^2} + [x + 1][x - 1]\] ;
c] \[[2x - 1][2x + 1] - {[2x - 3]^2}\] ;
d] \[{[x + 2]^3} - {[x + 1]^3} - {[x + 1]^2}\].
Lời giải chi tiết
\[\eqalign{ & a]\,\,\left[ {x + 3} \right]\left[ {x - 2} \right] + {\left[ {x - 3} \right]^2} \cr & \,\,\,\,\, = {x^2} - 2x + 3x - 6 + {x^2} - 6x + 9 \cr & \,\,\,\,\, = 2{x^2} - 5x + 3 \cr & b]\,\,{\left[ {x + 2} \right]^2} + \left[ {x + 1} \right]\left[ {x - 1} \right] \cr & \,\,\,\,\, = {x^2} + 4x + 4 + {x^2} - 1 \cr & \,\,\,\,\, = 2{x^2} + 4x + 3 \cr & c]\,\,\left[ {2x - 1} \right]\left[ {2x + 1} \right] - {\left[ {2x - 3} \right]^2} \cr & \,\,\,\,\, = {\left[ {2x} \right]^2} - 1 - \left[ {4{x^2} - 12x + 9} \right] \cr & \,\,\,\,\, = 4{x^2} - 1 - 4{x^2} + 12x - 9 \cr & \,\,\,\,\, = 12x - 10 \cr & d]\,\,{\left[ {x + 2} \right]^3} - {\left[ {x + 1} \right]^3} - {\left[ {x + 1} \right]^2} \cr & \,\,\,\,\, = {x^3} + 6{x^2} + 12x + 8 - \left[ {{x^3} + 3{x^2} + 3x + 1} \right] - \left[ {{x^2} + 2x + 1} \right] \cr & \,\,\,\,\, = {x^3} + 6{x^2} + 12x + 8 - {x^3} - 3{x^2} - 3x - 1 - {x^2} - 2x - 1 \cr & \,\,\,\,\, = 2{x^2} + 7x + 6 \cr} \]