Nội dung bài giảng
Tìm \(x)\ biết:
a. \(5x\left( {x - 1} \right) = x - 1\)
b. \(2\left( {x + 5} \right) - {x^2} - 5x = 0\)
Giải:
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. \(2\left( {x + 5} \right) - {x^2} - 5x = 0\)
\(\eqalign{ & \Rightarrow 2\left( {x + 5} \right) - \left( {{x^2} + 5x} \right) = 0 \Rightarrow 2\left( {x + 5} \right) - x\left( {x + 5} \right) = 0 \cr & \Rightarrow \left( {x + 5} \right)\left( {2 - x} \right) = 0 \Rightarrow \left[ {\matrix{ {x + 5 = 0} \cr {2 - x = 0} \cr } \Rightarrow \left[ {\matrix{ {x = - 5} \cr {x = 2} \cr } } \right.} \right. \cr} \)