Gíup mình câu 108 với Question Gíup mình câu 108 với in progress 0 Môn Toán Khoii Minh 5 years 2020-10-25T18:58:22+00:00 2020-10-25T18:58:22+00:00 1 Answers 82 views 0
Answers ( )
Đáp án:
a) $C=-\dfrac{3\sqrt[]{x}}{2(\sqrt[]{x}+2)}$
b) $x>16$
Giải thích các bước giải:
a) Điều kiện: $x>0$ và $x \neq 9$
$C=\Bigg(\dfrac{\sqrt[]{x}}{3+\sqrt[]{x}}+\dfrac{x+9}{9-x}\Bigg):\Bigg(\dfrac{3\sqrt[]{x}+1}{x-3\sqrt[]{x}}-\dfrac{1}{\sqrt[]{x}}\Bigg)$
$=\dfrac{\sqrt[]{x}(\sqrt[]{x}-3)-x-9}{(\sqrt[]{x}+3)(\sqrt[]{x}-3)}:\dfrac{3\sqrt[]{x}+1-(\sqrt[]{x}-3)}{\sqrt[]{x}(\sqrt[]{x}-3)}$
$=\dfrac{-3(\sqrt[]{x}+3)}{(\sqrt[]{x}+3)(\sqrt[]{x}-3)}.\dfrac{\sqrt[]{x}(\sqrt[]{x}-3)}{2(\sqrt[]{x}+2)}$
$=\dfrac{-3}{\sqrt[]{x}-3}.\dfrac{\sqrt[]{x}(\sqrt[]{x}-3)}{2(\sqrt[]{x}+2)}$
$=-\dfrac{3\sqrt[]{x}}{2(\sqrt[]{x}+2)}$
b) $C<-1 ↔ C+1<0$
$↔ -\dfrac{3\sqrt[]{x}}{2(\sqrt[]{x}+2)}+1<0$
$↔ \dfrac{4-\sqrt[]{x}}{2(\sqrt[]{x}+2)}<0$
Vì $\sqrt[]{x}≥0$ nên $\sqrt[]{x}+2>0$
$→ \dfrac{4-\sqrt[]{x}}{2(\sqrt[]{x}+2)}<0$
$↔ 4-\sqrt[]{x}<0$
$↔ \sqrt[]{x}>4$
$→ x>16$