Đề bài
Tìm x, biết:
a] \[{x^2} - 6x = 0\]
b] \[{x^3} - 25x = 0\] ;
c] \[{[2x - 5]^2} - x[2x - 5] = 0\]
d] \[{[3x - 1]^2} - {[x + 3]^2} = 0\]
e] \[{x^2} + 2x - 15 = 0\] .
Lời giải chi tiết
\[\eqalign{ & a]\,\,{x^2} - 6x = 0 \cr & \,\,\,\,x\left[ {x - 6} \right] = 0 \cr} \]
\[x = 0\] hoặc \[x - 6 = 0\]
\[x = 0\] hoặc \[x = 6\]
\[\eqalign{ & b]\,\,{x^3} - 25x = 0 \cr & \,x\left[ {{x^2} - 25} \right] = 0 \cr} \]
\[x = 0\] hoặc \[{x^2} - 25 = 0\]
\[x = 0\] hoặc \[x = \pm 5\]
\[\eqalign{ & c]\,\,{\left[ {2x - 5} \right]^2} - x\left[ {2x - 5} \right] = 0 \cr & \,\,\,\,\,\,\,\,\left[ {2x - 5} \right]\left[ {2x - 5 - x} \right] = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {2x - 5} \right]\left[ {x - 5} \right] = 0 \cr} \]
\[2x - 5 = 0\] hoặc \[x - 5 = 0\]
\[x = {5 \over 2}\] hoặc \[x = 5\]
\[\eqalign{ & d]\,\,{\left[ {3x - 1} \right]^2} - {\left[ {x + 3} \right]^2} = 0 \cr & \left[ {\left[ {3x - 1} \right] - \left[ {x + 3} \right]} \right]\left[ {\left[ {3x - 1} \right] + \left[ {x + 3} \right]} \right] = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {3x - 1 - x - 3} \right]\left[ {3x - 1 + x + 3} \right] = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {2x - 4} \right]\left[ {4x + 2} \right] = 0 \cr} \]
\[2x - 4 = 0\] hoặc \[4x + 2 = 0\]
\[x = 2\] hoặc \[x = - {1 \over 2}\]
\[\eqalign{ & e]\,\,{x^2} + 2x - 15 = 0 \cr & \,\,\,\,\,\,\,\,\,{x^2} + 2x + 1 - 16 = 0 \cr & \,\,\,\,\,\left[ {{x^2} + 2x + 1} \right] - 16 = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left[ {x + 1} \right]^2} - {4^2} = 0 \cr & \left[ {x + 1 - 4} \right]\left[ {x + 1 + 4} \right] = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {x - 3} \right]\left[ {x + 5} \right] = 0 \cr} \]
\[x - 3 = 0\] hoặc \[x + 5 = 0\]
\[x = 3\] hoặc \[x = - 5\]