toán 8

H

hien_vuthithanh

$x^2+11x+5$=$x^2$+2.x.$\dfrac{11}{2}$+$\dfrac{121}{4}$-$\dfrac{101}{4}$=$(x+\dfrac{11}{2})^2$-$\dfrac{101}{4}$ \geq $\dfrac{-101}{4}$
\Rightarrow min= $\dfrac{-101}{4}$ \Leftrightarrow x=$\dfrac{-11}{2}$

câu 2 này chỉ có min thôi k có max

nếu là min thì làm như sau
$x^2+5x-11$=$(x+\dfrac{5}{2})^2$-$\dfrac{69}{4}$ \geq $\dfrac{-69}{4}$
\Rightarrow min=$\dfrac{-69}{4}$ tại x=$\dfrac{-5}{2}$
 
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