1. Cho B= ($\frac{1}{√x+3}$ + $\frac{3}{x√x -9√x}$):( $\frac{√x}{√x+3}$ – $\frac{3√x-3}{x+3√x}$)
a) Rút Gọn B
b) Tìm X để B>1
2. Cho C= $\frac{x+2}{x√x+1}$ + $\frac{√x-1}{x+√x+1}$ – $\frac{1}{√x+1}$
a) Rút Gọn C
b) CHứng minh Rằng C luôn có giá trị không âm x
3. Cho D= ( $\frac{2√x+x}{x√x-1}$ – $\frac{1}{√x-1}$) : $\frac{x-1}{x+√x+1}$
a) Rút Gọn D
b) Tìm x để x<$\frac{-1}{2}$
MỌI NGƯỜI GIÚP MÌNH VỚI Ạk
Đáp án:
$\begin{array}{l}
1)a)Dkxd:x \ge 0;x \ne 9\\
B = \left( {\dfrac{1}{{\sqrt x + 3}} + \dfrac{3}{{x\sqrt x – 9\sqrt x }}} \right):\left( {\dfrac{{\sqrt x }}{{\sqrt x + 3}} – \dfrac{{3\sqrt x – 3}}{{x + 3\sqrt x }}} \right)\\
= \left( {\dfrac{1}{{\sqrt x + 3}} + \dfrac{3}{{\sqrt x \left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}} \right)\\
:\dfrac{{\sqrt x .\sqrt x – 3\sqrt x + 3}}{{\sqrt x .\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x – 3} \right) + 3}}{{\sqrt x \left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x + 3} \right)}}{{x – 3\sqrt x + 3}}\\
= \dfrac{{x – 3\sqrt x + 3}}{{\sqrt x – 3}}.\dfrac{1}{{x – 3\sqrt x + 3}}\\
= \dfrac{1}{{\sqrt x – 3}}\\
b)B > 1\\
\Rightarrow \dfrac{1}{{\sqrt x – 3}} – 1 > 0\\
\Rightarrow \dfrac{{1 – \sqrt x + 3}}{{\sqrt x – 3}} > 0\\
\Rightarrow \dfrac{{\sqrt x – 4}}{{\sqrt x – 3}} < 0\\
\Rightarrow 3 < \sqrt x < 4\\
\Rightarrow 9 < x < 16\\
2)a)Dkxd:x \ge 0\\
C = \dfrac{{x + 2}}{{x\sqrt x + 1}} + \dfrac{{\sqrt x – 1}}{{x + \sqrt x + 1}} – \dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{{x + 2 + \left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right) – x – \sqrt x – 1}}{{\left( {\sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x + 2 + x – 1 – x – \sqrt x – 1}}{{\left( {\sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x – \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x – 1} \right)}}{{x\sqrt x + 1}}\\
b)??\\
3)a)Dkxd:x \ge 0;x \ne 1\\
D = \left( {\dfrac{{2\sqrt x + x}}{{x\sqrt x – 1}} – \dfrac{1}{{\sqrt x – 1}}} \right):\dfrac{{x – 1}}{{x + \sqrt x + 1}}\\
= \dfrac{{2\sqrt x + x – \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{x – 1}}\\
= \dfrac{{2\sqrt x + x – x – \sqrt x – 1}}{{\sqrt x – 1}}.\dfrac{1}{{x – 1}}\\
= \dfrac{{\sqrt x – 1}}{{\sqrt x – 1}}.\dfrac{1}{{x – 1}}\\
= \dfrac{1}{{x – 1}}\\
b)D < – \dfrac{1}{2}\\
\Rightarrow \dfrac{1}{{x – 1}} < – \dfrac{1}{2}\\
\Rightarrow \dfrac{1}{{x – 1}} + \dfrac{1}{2} < 0\\
\Rightarrow \dfrac{{2 + x – 1}}{{2\left( {x – 1} \right)}} < 0\\
\Rightarrow \dfrac{{x + 1}}{{2\left( {x – 1} \right)}} < 0\\
\Rightarrow x – 1 < 0\left( {do:x \ge 0} \right)\\
\Rightarrow x < 1\\
Vay\,0 \le x < 1
\end{array}$
(B2 câu b ko thể luôn có giá trị ko âm với mọi x được.)