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Tìm \(x)\ biết:
a. \(5x\left( {x – 1} \right) = x – 1\)
b. \(2\left( {x + 5} \right) – {x^2} – 5x = 0\)
a. \(5x\left( {x – 1} \right) = x – 1\)
\(\eqalign{ & \Rightarrow 5x\left( {x – 1} \right) – \left( {x – 1} \right) = 0 \Rightarrow \left( {x – 1} \right)\left( {5x – 1} \right) = 0 \cr & \Rightarrow \left[ {\matrix{ {x – 1 = 0} \cr {5x – 1 = 0} \cr } \Rightarrow \left[ {\matrix{ {x = 1} \cr {x = {1 \over 5}} \cr } } \right.} \right. \cr} \)
b. \(2\left( {x + 5} \right) – {x^2} – 5x = 0\)
\(\eqalign{ & \Rightarrow 2\left( {x + 5} \right) – \left( {{x^2} + 5x} \right) = 0 \Rightarrow 2\left( {x + 5} \right) – x\left( {x + 5} \right) = 0 \cr & \Rightarrow \left( {x + 5} \right)\left( {2 – x} \right) = 0 \Rightarrow \left[ {\matrix{ {x + 5 = 0} \cr {2 – x = 0} \cr } \Rightarrow \left[ {\matrix{ {x = – 5} \cr {x = 2} \cr } } \right.} \right. \cr} \)