Đáp án:
Tham khảo
Giải thích các bước giải:
$P=(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}):\dfrac{\sqrt{x}}{x+\sqrt{x}}$
a) $ĐK:x≥0$
$=(\dfrac{\sqrt{x}+1+x}{\sqrt{x}(\sqrt{x}+1)}).\dfrac{x+\sqrt{x}}{\sqrt{x}}$
$=\dfrac{\sqrt{x}+1+x}{\sqrt{x}(\sqrt{x}+1)}.\dfrac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}}$
$=\dfrac{\sqrt{x}+1+x}{\sqrt{x}}$
b)Ta có:x=4
$⇔\dfrac{\sqrt{4}+1+4}{\sqrt{4}}=\dfrac{2+1+4}{2}=\dfrac{7}{2}$
c)Để $P=\dfrac{13}{3}⇔\dfrac{\sqrt{x}+1+x}{\sqrt{x}}\dfrac{13}{3}$
$⇔3(\sqrt{x}+1+x)=13\sqrt{x}$
$⇔3\sqrt{x}+3+3x-13\sqrt{x}=0$
$⇔3x-10\sqrt{x}+3=0$
$⇔3(\sqrt{x})^2-9\sqrt{x}-\sqrt{x}+3=0$
$⇔3\sqrt{x}(\sqrt{x}-3)-(\sqrt{x}-3)=0$
$⇔(\sqrt{x}-3)(3\sqrt{x}-1)=0$
⇔\(\left[ \begin{array}{l}\sqrt{x}-3=0\\3\sqrt{x}-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}\sqrt{x}=3\\\sqrt{x}=\dfrac{1}{3}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=9\\x=\dfrac{1}{9}\end{array} \right.\)
Vậy $x=9$và $x=\dfrac{1}{9}$thì $P=\dfrac{13}{3}$
