Đề bài
Tìm x, biết:
a] \[{[2x + 1]^2} - {[3 - 2x]^2} + 4 = 0\]
b] \[{[x - 1]^3} + [2 - x][4 + 2x + {x^2}] + 3x[x + 2] = 17\]
c] \[[x + 2][{x^2} - 2x + 4] - x[{x^2} - 2] = 15\]
Lời giải chi tiết
\[\eqalign{ & a]\,\,{\left[ {2x + 1} \right]^2} - {\left[ {3 - 2x} \right]^2} + 4 = 0 \cr & \Leftrightarrow\left[ {4{x^2} + 4x + 1} \right] - \left[ {9 - 12x + 4{x^2}} \right] + 4 = 0 \cr & \Leftrightarrow 4{x^2} + 4x + 1 - 9 + 12x - 4{x^2} + 4 = 0 \cr & \Leftrightarrow 16x - 4 = 0 \cr & \Leftrightarrow 16x = 4 \cr & \Leftrightarrowx = {1 \over 4} \cr & b]\,\,{\left[ {x - 1} \right]^3} + \left[ {2 - x} \right]\left[ {4 + 2x + {x^2}} \right] + 3x\left[ {x + 2} \right] = 17 \cr & \Leftrightarrow{x^3} - 3{x^2} + 3x - 1 + 8 - {x^3} + 3{x^2} + 6x = 17 \cr & \Leftrightarrow9x + 7 = 17 \cr & \Leftrightarrow9x = 10 \cr & \Leftrightarrowx = {{10} \over 9} \cr & c]\,\,\left[ {x + 2} \right]\left[ {{x^2} - 2x + 4} \right] - x\left[ {{x^2} - 2} \right] = 15 \cr & \Leftrightarrow{x^3} + 8 - {x^3} + 2x = 1 \cr &\Leftrightarrow - 8 + 2x = 15 \cr & \Leftrightarrow2x = 23 \cr & \Leftrightarrowx = {{23} \over 2} \cr} \]