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 \(x = - 5\)
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