Toán Lớp 8: B=x mũ 2+5x+1
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Toán Lớp 8: B=x mũ 2+5x+1
Tìm giá trị nhỏ nhất
Comments ( 2 )
$B=x^2+2.x.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{21}{4}$
$B=(x+\dfrac{5}{2})^2-\dfrac{21}{4}$
$(x+\dfrac{5}{2})^2≥0∀x$
$⇒(x+\dfrac{5}{2})^2-\dfrac{21}{4}≥\dfrac{-21}{4}$
Dấu $”=”$ xảy ra khi
$⇔x=\dfrac{-5}{2}$
Vậy $B_{min}=\dfrac{-21}{4}⇔x=\dfrac{-5}{2}$