Tìm x, biết: a) 4x ( x-3 ) + 6 ( 3-x ) =0
a) \(4x\left( {x - 3} \right) + 6\left( {3 - x} \right) = 0\) \(\left( {x - 3} \right)\left( {4x - 6} \right) = 0\) | b) \({x^3} - x\left( {x - 1} \right)\left( {x + 1} \right) = 14\) \({x^3} - x\left( {{x^2} - 1} \right) = 14\) | |
Trường hợp 1: \(x - 3 = 0\) \(x = 3\). Vậy \(x \in \left\{ {3;\frac{3}{2}} \right\}\). | Trường hợp 2: \(4x - 6 = 0\) \(4x = 6\) \(x = \frac{3}{2}\). | \({x^3} - {x^3} + x = 14\) \(x = 14\) Vậy \(x = 14\). |
c) \({\left( {{x^2} - x} \right)^2} + 2\left( {{x^2} - x} \right) = 8\)
\({\left( {{x^2} - x} \right)^2} + 2\left( {{x^2} - x} \right) - 8 = 0\)
\({\left( {{x^2} - x} \right)^2} + 2\left( {{x^2} - x} \right) + 1 - 9 = 0\)
\({\left[ {\left( {{x^2} - x} \right) + 1} \right]^2} - {3^2} = 0\)
\(\left( {{x^2} - x + 1 + 3} \right)\left( {{x^2} - x + 1 - 3} \right) = 0\)
\(\left( {{x^2} - x + 4} \right)\left( {{x^2} - x - 2} \right) = 0\)
\(\left[ {{{\left( {x - \frac{1}{2}} \right)}^2} + \frac{{15}}{4}} \right].\left( {{x^2} + x - 2x - 2} \right) = 0\)
Suy ra \({x^2} + x - 2x - 2 = 0\) (Vì \({\left( {x - \frac{1}{2}} \right)^2} + \frac{{15}}{4} > 0\))
\(x\left( {x + 1} \right) - 2\left( {x + 1} \right) = 0\)
\(\left( {x + 1} \right)\left( {x - 2} \right) = 0\)
Suy ra \(x + 1 = 0\) hoặc \(x - 2 = 0\)
Do đó \(x = - 1\) hoặc \(x = 2.\)
Vậy \(x \in \left\{ { - 1;2} \right\}\).