Tìm \(x,\) biết
a. \({x^2} - 2x - 3 = 0\)
b. \(2{x^2} + 5x - 3 = 0\)
a. \({x^2} - 2x - 3 = 0\)
\(\eqalign{ & \Rightarrow {x^2} - 2x + 1 - 4 = 0 \Rightarrow {\left( {x - 1} \right)^2} - {2^2} = 0 \cr & \Rightarrow \left( {x - 1 + 2} \right)\left( {x - 1 - 2} \right) = 0 \Rightarrow \left( {x + 1} \right)\left( {x - 3} \right) \cr} \)
\( \Rightarrow x + 1 = 0\) hoặc \(x - 3 = 0\)
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\(\eqalign{ & x + 1 = 0 \Rightarrow x = - 1 \cr & x - 3 = 0 \Rightarrow x = 3 \cr} \)
Vậy \(x = - 1\)và \(x = 3\)
b. \(2{x^2} + 5x - 3 = 0\)
\(\eqalign{ & \Rightarrow 2{x^2} + 6x - x - 3 = 0 \Rightarrow 2x\left( {x + 3} \right) - \left( {x + 3} \right) = 0 \cr & \Rightarrow \left( {x + 3} \right)\left( {2x - 1} \right) = 0 \cr} \) \( \Rightarrow x + 3 = 0\) hoặc \(2x - 1 = 0\)
\(\eqalign{ & x + 3 = 0 \Rightarrow x = - 3 \cr & 2x - 1 = 0 \Rightarrow x = {1 \over 2} \cr} \)
Vậy \(x = - 3\) hoặc \(x = {1 \over 2}\)