Toán Lớp 8: cho A=4x^2+4x+2 b=2x^2-2x+1 c=-15-x^2+6x b,c/m c luôn âm c, tìm gtln (gtnn) của a,b,c
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Toán Lớp 8: cho A=4x^2+4x+2 b=2x^2-2x+1 c=-15-x^2+6x b,c/m c luôn âm c, tìm gtln (gtnn) của a,b,c
Comments ( 2 )
b)dpcm\\
c)MinA = 1\\
MinB = \dfrac{1}{2}\\
MaxC = – 6
\end{array}\)
b)C = – 15 – {x^2} + 6x\\
= – \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 dpcm\\
c)A = 4{x^2} + 4x + 2\\
= 4{x^2} + 4x + 1 + 1\\
= {\left( {2x + 1} \right)^2} + 1\\
Do:{\left( {2x + 1} \right)^2} \ge 0\forall x\\
\to {\left( {2x + 1} \right)^2} + 1 \ge 1\\
\to MinA = 1\\
\Leftrightarrow x = – \dfrac{1}{2}\\
B = 2{x^2} – 2x + 1\\
= 2{x^2} – 2.x\sqrt 2 .\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} + \dfrac{1}{2}\\
= {\left( {x\sqrt 2 – \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{1}{2}\\
Do:{\left( {x\sqrt 2 – \dfrac{1}{{\sqrt 2 }}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt 2 – \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{1}{2} \ge \dfrac{1}{2}\\
\to MinB = \dfrac{1}{2}\\
\Leftrightarrow x\sqrt 2 – \dfrac{1}{{\sqrt 2 }} = 0\\
\to x = \dfrac{1}{2}\\
C = – {\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 \le – 6\\
\to MaxC = – 6\\
\Leftrightarrow x = 3
\end{array}\)