Toán Lớp 8: a, $x^{2}$ – 3x + 5 ≥ 0
b, $x^{2}$ + 5x – 3
c, – $x^{2}$ + 7x + 3
d, $4x^{2}$ – 5x – 1 ≥ 0
* Câu b ≤ 0, câu c ≥ 0. Vì 1 số trục trặc nên mình không viết được ạ
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Toán Lớp 8: a, $x^{2}$ – 3x + 5 ≥ 0
b, $x^{2}$ + 5x – 3
c, – $x^{2}$ + 7x + 3
d, $4x^{2}$ – 5x – 1 ≥ 0
* Câu b ≤ 0, câu c ≥ 0. Vì 1 số trục trặc nên mình không viết được ạ
Comments ( 1 )
a){x^2} – 3x + 5 \ge 0\\
\Leftrightarrow {x^2} – 2.x.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{{11}}{4} \ge 0\\
\Leftrightarrow {\left( {x – \dfrac{3}{2}} \right)^2} + \dfrac{{11}}{4} \ge 0\left( {tmdk} \right)\\
Vậy\,x \in R\\
b){x^2} + 5x – 3 \le 0\\
\Leftrightarrow {x^2} + 2.x.\dfrac{5}{2} + \dfrac{{25}}{4} – \dfrac{{13}}{4} \le 0\\
\Leftrightarrow {\left( {x + \dfrac{5}{2}} \right)^2} \le \dfrac{{13}}{4}\\
\Leftrightarrow \dfrac{{ – \sqrt {13} – 5}}{2} \le x \le \dfrac{{\sqrt {13} – 5}}{2}\\
Vậy\,\dfrac{{ – \sqrt {13} – 5}}{2} \le x \le \dfrac{{\sqrt {13} – 5}}{2}\\
c) – {x^2} + 7x + 3 \ge 0\\
\Leftrightarrow {x^2} – 7x – 3 \le 0\\
\Leftrightarrow {x^2} – 2.x.\dfrac{7}{2} + \dfrac{{49}}{4} – \dfrac{{61}}{4} \le 0\\
\Leftrightarrow {\left( {x – \dfrac{7}{2}} \right)^2} \le \dfrac{{61}}{4}\\
\Leftrightarrow \dfrac{{7 – \sqrt {61} }}{2} \le x \le \dfrac{{7 + \sqrt {61} }}{2}\\
Vậy\,\dfrac{{7 – \sqrt {61} }}{2} \le x \le \dfrac{{7 + \sqrt {61} }}{2}\\
d)4{x^2} – 5x – 1 \ge 0\\
\Leftrightarrow 4{x^2} – 2.2x.\dfrac{5}{4} + \dfrac{{25}}{{16}} – \dfrac{{41}}{{16}} \ge 0\\
\Leftrightarrow {\left( {2x – \dfrac{5}{4}} \right)^2} \ge \dfrac{{41}}{{16}}\\
\Leftrightarrow \left[ \begin{array}{l}
2x – \dfrac{5}{4} \ge \dfrac{{\sqrt {41} }}{4}\\
2x – \dfrac{5}{4} \le \dfrac{{ – \sqrt {41} }}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge \dfrac{{5 + \sqrt {41} }}{8}\\
x \le \dfrac{{5 – \sqrt {41} }}{8}
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x \ge \dfrac{{5 + \sqrt {41} }}{8}\\
x \le \dfrac{{5 – \sqrt {41} }}{8}
\end{array} \right.
\end{array}$