Toán Lớp 8: tìm giá trị nhỏ nhất E= 3x^2+10x+8
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Toán Lớp 8: tìm giá trị nhỏ nhất E= 3x^2+10x+8
Comments ( 1 )
E = 3{x^2} + 10x + 8\\
= 3.\left( {{x^2} + \dfrac{{10}}{3}.x} \right) + 8\\
= 3.\left( {{x^2} + 2.x.\dfrac{5}{3} + \dfrac{{25}}{9}} \right) – 3.\dfrac{{25}}{9} + 8\\
= 3.{\left( {x + \dfrac{5}{3}} \right)^2} – \dfrac{1}{3}\\
Do:{\left( {x + \dfrac{5}{3}} \right)^2} \ge 0\\
\Leftrightarrow 3.{\left( {x + \dfrac{5}{3}} \right)^2} \ge 0\\
\Leftrightarrow 3.{\left( {x + \dfrac{5}{3}} \right)^2} – \dfrac{1}{3} \ge – \dfrac{1}{3}\\
\Leftrightarrow E \ge – \dfrac{1}{3}\\
\Leftrightarrow GTNN:E = – \dfrac{1}{3}\,khi:x = – \dfrac{5}{3}
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