Tìm x , biết: a) 3 ( x − 1 )^2 − 3x ( x − 5 ) = 0. b) ( x + 3 )^2 − 5x − 15 = 0.
Hướng dẫn giải
a) \(3{\left( {x - 1} \right)^2} - 3x\left( {x - 5} \right) = 0\) \(3\left( {{x^2} - 2x + 1} \right) - \left( {3{x^2} - 15x} \right) = 0\) \(3{x^2} - 6x + 3 - 3{x^2} + 15x = 0\) \(\left( {3{x^2} - 3{x^2}} \right) + \left( { - 6x + 15x} \right) + 3 = 0\) \(9x = - 3\) \(x = - \frac{1}{3}.\) Vậy \(x = - \frac{1}{3}.\)
| b) \[{\left( {x + 3} \right)^2} - 5x - 15 = 0\] \[{\left( {x + 3} \right)^2} - \left( {5x - 15} \right) = 0\] \[{\left( {x + 3} \right)^2} - 5\left( {x + 3} \right) = 0\] \[\left( {x + 3} \right)\left( {x + 3 - 5} \right) = 0\] \[\left( {x + 3} \right)\left( {x - 2} \right) = 0\] Suy ra \[x + 3 = 0\] hoặc \[x - 2 = 0\] \[x = - 3\] hoặc \[x = 2\]. Vậy \[x \in \left\{ { - 3\,;\,\,2} \right\}\]. |
c) \[2{x^5} - 4{x^3} + 2x = 0\]
\[2x\left( {{x^4} - 2{x^2} + 1} \right) = 0\]
\[2x\left[ {{{\left( {{x^2}} \right)}^2} - \,\,2{x^2} + {1^2}} \right] = 0\]
\[2x{\left( {{x^2} - 1} \right)^2} = 0\]
\[2x{\left[ {\left( {x - 1} \right)\left( {x + 1} \right)} \right]^2} = 0\]
\[2x{\left( {x - 1} \right)^2}{\left( {x + 1} \right)^2} = 0\]
Suy ra \[2x = 0\] hoặc \[{\left( {x - 1} \right)^2} = 0\] hoặc \[{\left( {x + 1} \right)^2} = 0\]
\[x = 0\] hoặc \[x - 1 = 0\] hoặc \[x + 1 = 0\]
\[x = 0\] hoặc \[x = 1\] hoặc \[x = - 1\]
Vậy \[x \in \left\{ { - 1\,;\,\,0\,;\,\,1} \right\}\].