Toán Lớp 8: chứng minh các đã thức sau luôn âm với mọi x
-x2+6x -15
-9×2+24-18
(x-3).(1-X)-2
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Toán Lớp 8: chứng minh các đã thức sau luôn âm với mọi x
-x2+6x -15
-9×2+24-18
(x-3).(1-X)-2
Comments ( 1 )
a) – {x^2} + 6x – 15\\
= – \left( {{x^2} – 6x + 15} \right)\\
= – \left( {{x^2} – 6x + 9 + 6} \right)\\
= – {\left( {x – 3} \right)^2} – 6\\
Do:{\left( {x – 3} \right)^2} \ge 0\forall x\\
\to – {\left( {x – 3} \right)^2} \le 0\\
\to – {\left( {x – 3} \right)^2} – 6 < 0\\
\to – {x^2} + 6x – 15 < 0\\
b) – 9{x^2} + 24x – 18\\
= – \left( {9{x^2} – 24x + 18} \right)\\
= – \left( {9{x^2} – 2.3x.4 + 16 + 2} \right)\\
= – {\left( {3x – 4} \right)^2} – 2\\
Do:{\left( {3x – 4} \right)^2} \ge 0\forall x \ge 0\\
\to – {\left( {3x – 4} \right)^2} \le 0\\
\to – {\left( {3x – 4} \right)^2} – 2 < 0\\
\to – 9{x^2} + 24x – 18 < 0\\
c)\left( {x – 3} \right)\left( {1 – x} \right) – 2\\
= – {x^2} + 4x – 3 – 2\\
= – \left( {{x^2} – 4x + 5} \right)\\
= – \left( {{x^2} – 4x + 4 + 1} \right)\\
= – {\left( {x – 2} \right)^2} – 1\\
Do:{\left( {x – 2} \right)^2} \ge 0\forall x\\
\to – {\left( {x – 2} \right)^2} \le 0\\
\to – {\left( {x – 2} \right)^2} – 1 < 0\\
\to \left( {x – 3} \right)\left( {1 – x} \right) – 2 < 0
\end{array}\)