Đáp án:
a) $A=29$
b) $B=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}$
c) ${{\min }_{\text{P}}}=4\Leftrightarrow x=1$
Giải thích các bước giải:
$A=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}-3}$ và $B=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}$ với $x\ge 0,x\ne 4,x\ne 9$
a)Với $x=16$
$A=\dfrac{16+2\sqrt{16}+5}{\sqrt{16}-3}=\dfrac{16+2.4+5}{4-3}=29$
b)
$B=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}$
$B=\dfrac{2\sqrt{x}-9}{\left( \sqrt{x}-2 \right)\left( \sqrt{x}-3 \right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}$
$B=\dfrac{2\sqrt{x}-9-\left( \sqrt{x}+3 \right)\left( \sqrt{x}-3 \right)+\left( 2\sqrt{x}+1 \right)\left( \sqrt{x}-2 \right)}{\left( \sqrt{x}-2 \right)\left( \sqrt{x}-3 \right)}$
$B=\dfrac{2\sqrt{x}-9-\left( x-9 \right)+\left( 2x-4\sqrt{x}+\sqrt{x}-2 \right)}{\left( \sqrt{x}-2 \right)\left( \sqrt{x}-3 \right)}$
$B=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left( \sqrt{x}-2 \right)\left( \sqrt{x}-3 \right)}$
$B=\dfrac{x-\sqrt{x}-2}{\left( \sqrt{x}-2 \right)\left( \sqrt{x}-3 \right)}$
$B=\dfrac{\left( \sqrt{x}-2 \right)\left( \sqrt{x}+1 \right)}{\left( \sqrt{x}-2 \right)\left( \sqrt{x}-3 \right)}$
$B=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}$
c)
$P=A:B$
$P=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}-3}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}$
$P=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}+1}$
$P=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}+1}$
$P=\dfrac{{{\left( \sqrt{x}+1 \right)}^{2}}}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}+1}$
$P=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}$
$P\ge 2\sqrt{\left( \sqrt{x}+1 \right)\cdot \dfrac{4}{\sqrt{x}+1}}$
$P\ge 2\sqrt{4}$
$P\ge 4$
Dấu “=” xảy ra khi $\sqrt{x}+1=\dfrac{4}{\sqrt{x}+1}\Leftrightarrow x=1$
Vậy ${{\min }_{\text{P}}}=4\Leftrightarrow x=1$
