Giải các pt sau:
a, ($2x^{2}$ -3) – 4$(x-1)^{2}$ =0
b, $x^{3}$+ $3x^{2}$ +x+3=0
c,$x^{3}$ – $6x^{2}$ + 11x – 6 = 0
Giải các pt sau:
a, ($2x^{2}$ -3) – 4$(x-1)^{2}$ =0
b, $x^{3}$+ $3x^{2}$ +x+3=0
c,$x^{3}$ – $6x^{2}$ + 11x – 6 = 0
Đáp án:
c) \(\left[ \begin{array}{l}
x = 1\\
x = 2\\
x = 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)2{x^2} – 3 – 4\left( {{x^2} – 2x + 1} \right) = 0\\
\to 2{x^2} – 3 – 4{x^2} + 8x – 4 = 0\\
\to – 2{x^2} + 8x – 7 = 0\\
\Delta ‘ = 16 – 2.7 = 2\\
\to \left[ \begin{array}{l}
x = \dfrac{{4 + \sqrt 2 }}{2}\\
x = \dfrac{{4 – \sqrt 2 }}{2}
\end{array} \right.\\
b){x^2}\left( {x + 3} \right) + \left( {x + 3} \right) = 0\\
\to \left( {x + 3} \right)\left( {{x^2} + 1} \right) = 0\\
\to x + 3 = 0\left( {do:{x^2} + 1 > 0\forall x} \right)\\
\to x = – 3\\
c){x^3} – {x^2} – 5{x^2} + 5x + 6x – 6 = 0\\
\to {x^2}\left( {x – 1} \right) – 5x\left( {x – 1} \right) + 6\left( {x – 1} \right) = 0\\
\to \left( {x – 1} \right)\left( {{x^2} – 5x + 6} \right) = 0\\
\to \left( {x – 1} \right)\left( {x – 2} \right)\left( {x – 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = 2\\
x = 3
\end{array} \right.
\end{array}\)