Toán Lớp 9: giải các phương trình sau
1, x ≤x ²
2.6x ²-x-2
3,x ²-x+2 ≥0
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Toán Lớp 9: giải các phương trình sau
1, x ≤x ²
2.6x ²-x-2
3,x ²-x+2 ≥0
Comments ( 1 )
1)x \le {x^2}\\
\Leftrightarrow {x^2} – x \ge 0\\
\Leftrightarrow x.\left( {x – 1} \right) \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
x – 1 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
x – 1 \le 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
x \ge 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
x \le 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le 0
\end{array} \right.\\
Vậy\,x \le 0\,hoặc\,x \ge 1\\
2)6{x^2} – x – 2 \le 0\\
\Leftrightarrow 6{x^2} – 4x + 3x – 2 \le 0\\
\Leftrightarrow \left( {3x – 2} \right)\left( {2x + 1} \right) \le 0\\
\Leftrightarrow – \dfrac{1}{2} \le x \le \dfrac{2}{3}\\
Vậy\, – \dfrac{1}{2} \le x \le \dfrac{2}{3}\\
3){x^2} – x + 2 \ge 0\\
\Leftrightarrow {x^2} – 2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{7}{4} \ge 0\\
\Leftrightarrow {\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{7}{4} \ge 0\left( {tm} \right)\\
Vậy\,x \in \emptyset
\end{array}$