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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: Tìm GTNN: $\frac{x-19}{\sqrt{x}+9}$ x$\geq$ 0,x$\neq$ 4

Toán Lớp 9: Tìm GTNN:
$\frac{x-19}{\sqrt{x}+9}$
x$\geq$ 0,x$\neq$ 4

Comments ( 1 )

  1. Giải đáp:
    $min_A= -\dfrac{19}{9}  \Leftrightarrow x=0.$
    Lời giải và giải thích chi tiết:
    $A=\dfrac{x-19}{x+9}=\dfrac{x+9-28}{x+9}=1-\dfrac{28}{x+9}\\ x+9 \ge 9 \ \forall \ x \ge 0;x \ne 4\\ \Rightarrow \dfrac{28}{x+9} \le \dfrac{28}{9} \ \forall \ x \ge 0;x \ne 4\\ \Rightarrow -\dfrac{28}{x+9} \ge -\dfrac{28}{9} \ \forall \ x \ge 0;x \ne 4\\ \Rightarrow 1-\dfrac{28}{x+9} \ge -\dfrac{19}{9} \ \forall \ x \ge 0;x \ne 4\\ \Leftrightarrow A \ge -\dfrac{19}{9} \ \forall \ x \ge 0;x \ne 4$
    Dấu “=” xảy ra $\Leftrightarrow x=0$
    Vậy $min_A= -\dfrac{19}{9}  \Leftrightarrow x=0.$

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222-9+11+12:2*14+14 = ? ( )