Đề bài
Tìm x, biết:
a) \(5x(x - 4) - {x^2} + 16 = 0\) ;
b) \(x + 6{x^2} + 9{x^3} = 0\) ;
c) \({x^2} - 4x + 3 = 0\) .
Lời giải chi tiết
\(\eqalign{ & a)\,\,5x\left( {x - 4} \right) - {x^2} + 16 = 0 \cr & \,\,\,\,\,\,\,\,\,5x\left( {x - 4} \right) - \left( {{x^2} - 16} \right) = 0 \cr & 5x\left( {x - 4} \right) - \left( {x - 4} \right)\left( {x + 4} \right) = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x - 4} \right)\left( {5x - x - 4} \right) = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x - 4} \right)\left( {4x - 4} \right) = 0 \cr} \)
\(x - 4 = 0\) hoặc \(4x - 4 = 0\)
\(x = 4\) hoặc \(x = 1\)
\(\eqalign{ & b)\,\,x + 6{x^2} + 9{x^3} = 0 \cr & x\left( {1 + 6x + 9{x^2}} \right) = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,x{\left( {3x + 1} \right)^2} = 0 \cr} \)
\(x = 0\) hoặc \({\left( {3x + 1} \right)^2} = 0\)
\(x = 0\) hoặc \(3x + 1 = 0\)
\(x = 0\) hoặc \(x = - {1 \over 3}\)
\(\eqalign{ & c)\,\,{x^2} - 4x + 3 = 0 \cr & \,\,\,\,\,\,\,\,{x^2} - x - 3x + 3 = 0 \cr & x\left( {x - 1} \right) - 3\left( {x - 1} \right) = 0 \cr & \,\,\,\,\,\,\,\,\,\,\left( {x - 1} \right)\left( {x - 3} \right) = 0 \cr} \)
\(x - 1 = 0\) hoặc \(x - 3 = 0\)
\(x = 1\) hoặc \(x = 3\)
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