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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 \(x – 3 = 0\)
\(\eqalign{ & x + 1 = 0 \Rightarrow x = – 1 \cr & x – 3 = 0 \Rightarrow x = 3 \cr} \)
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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 \(x = {1 \over 2}\)
Mục lục môn Toán 8 (SBT)