Bài tập 1:
a,Cho biểu thức A= ($\frac{3x + \sqrt[2]{9} – 3}{x + \sqrt[2]{x} – 2}$ + $\frac{1}{\sqrt[2]{x} -1}$ + $\frac{1}{\sqrt[2]{x} + 2}$) : $\frac{1}{x – 1}$
-, Tìm điều kiện xác định
-, Rút gọn A
b,Cho biểu thức C= ( $\frac{2x + 1}{x\sqrt[2]{x} – 1}$ – $\frac{\sqrt[2]{x}}{x+\sqrt[2]{x}+1}$)*( $\frac{1+x\sqrt[2]{x}}{1+\sqrt[2]{x}}$ – $\sqrt[2]{x}$)
-, Tìm điều kiện xác định + rút gọn
-, Tìm x để B=5
Đáp án:
\[\begin{array}{l}
B1)a)\\
Dkxd:x \ge 0;x \ne 1\\
A = \left( {\dfrac{{3x + \sqrt {9x} – 3}}{{x + \sqrt x – 2}} + \dfrac{1}{{\sqrt x – 1}} + \dfrac{1}{{\sqrt x + 2}}} \right):\dfrac{1}{{x – 1}}\\
= \dfrac{{3x + 3\sqrt x – 3 + \left( {\sqrt x + 2} \right) + \sqrt x – 1}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 2} \right)}}.\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)\\
= \dfrac{{3x + 5\sqrt x – 2}}{{\sqrt x + 2}}.\left( {\sqrt x + 1} \right)\\
= \dfrac{{3\sqrt x – \sqrt x + 6\sqrt x – 2}}{{\sqrt x + 2}}.\left( {\sqrt x + 1} \right)\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {3\sqrt x – 1} \right)}}{{\sqrt x + 2}}.\left( {\sqrt x + 1} \right)\\
= \left( {3\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)\\
= 3x + 2\sqrt x – 1\\
b)Dkxd:x \ge 0;x \ne 1\\
C = \left( {\dfrac{{2x}}{{x\sqrt x – 1}} – \dfrac{{\sqrt x }}{{x + \sqrt x + 1}}} \right).\\
\left( {\dfrac{{1 + x\sqrt x }}{{1 + \sqrt x }} – \sqrt x } \right)\\
= \dfrac{{2x – \sqrt x \left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}.\left( {\dfrac{{\left( {1 + \sqrt x } \right)\left( {1 – \sqrt x + x} \right)}}{{1 + \sqrt x }} – \sqrt x } \right)\\
= \dfrac{{x + \sqrt x }}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}.{\left( {\sqrt x – 1} \right)^2}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}{{x + \sqrt x + 1}}\\
= \dfrac{{x – 1}}{{x + \sqrt x + 1}}\\
B = 5\\
\Rightarrow \dfrac{{x – 1}}{{x + \sqrt x + 1}} = 5\\
\Rightarrow x – 1 = 5x + 5\sqrt x + 5\\
\Rightarrow 4x + 5\sqrt x + 6 = 0\\
\Rightarrow x + \dfrac{5}{4}\sqrt x + \dfrac{3}{2} = 0\\
\Rightarrow x + 2.\sqrt x .\dfrac{5}{8} + \dfrac{{25}}{{64}} + \dfrac{{71}}{{64}} = 0\\
\Rightarrow {\left( {\sqrt x + \dfrac{5}{8}} \right)^2} + \dfrac{{71}}{{64}} = 0\left( {\text{vô}\,\text{nghiệm}} \right)
\end{array}\]
Vậy ko có x để B=5