Cho hai biểu thức $A=\dfrac{\sqrt[]{x}+4}{\sqrt[]{x}-1}$ và $B=\dfrac{3\sqrt[]{x}+1}{x+2\sqrt[]{x}-3}-\dfrac{2}{\sqrt[]{x}+3}$ (với $x\geq 0$, `xne 1

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1. $A =\dfrac{\sqrt x + 4}{\sqrt x -1}$

   $B =\dfrac{3\sqrt x+1}{x + 2\sqrt x -3} – \dfrac{2}{\sqrt x +3}$

   $(x\geqslant 0;\ x \ne 1)$

   $+)\quad x = 9\Rightarrow \sqrt x = 3$

   Thay vào $A$ ta được:

   $\quad A =\dfrac{3 +4}{3 -1}=\dfrac72$

   $+)\quad B = \dfrac{3\sqrt x +1}{\left(\sqrt x -1\right)\left(\sqrt x +3\right)}- \dfrac{2\left(\sqrt x – 1\right)}{\left(\sqrt x -1\right)\left(\sqrt x +3\right)}$

   $\to B =\dfrac{3\sqrt x + 1 – 2\sqrt x + 2}{\left(\sqrt x -1\right)\left(\sqrt x +3\right)}$

   $\to B = \dfrac{\sqrt x +3}{\left(\sqrt x -1\right)\left(\sqrt x +3\right)}$

   $\to B =\dfrac{1}{\sqrt x -1}\quad$ (đpcm)

   $+)\quad \dfrac{A}{B}\geqslant \dfrac x4 + 5$

   $\Leftrightarrow \dfrac{\dfrac{\sqrt x +4}{\sqrt x -1}}{\dfrac{1}{\sqrt x -1}}\geqslant \dfrac x4 + 5$

   $\Leftrightarrow \sqrt x + 4\geqslant \dfrac x4 + 5$

   $\Leftrightarrow \dfrac x4 – \sqrt x + 1 \geqslant 0$

   $\Leftrightarrow \left(\dfrac{\sqrt x}{2} – 1\right)^2 \geqslant 0$ (luôn đúng)

   Vậy $x \geqslant 0;\ x\ne 1$

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2. Đáp án:

   A=√x+4√x−1A=x+4x−1

   B=3√x+1x+2√x−3−2√x+3B=3x+1x+2x−3−2x+3

   (x⩾0; x≠1)(x⩾0; x≠1)

   +)x=9⇒√x=3+)x=9⇒x=3

   Thay vào AA ta được:

   A=3+43−1=72A=3+43−1=72

   +)B=3√x+1(√x−1)(√x+3)−2(√x−1)(√x−1)(√x+3)+)B=3x+1(x−1)(x+3)−2(x−1)(x−1)(x+3)

   →B=3√x+1−2√x+2(√x−1)(√x+3)→B=3x+1−2x+2(x−1)(x+3)

   →B=√x+3(√x−1)(√x+3)→B=x+3(x−1)(x+3)

   →B=1√x−1→B=1x−1 (đpcm)

   +)AB⩾x4+5+)AB⩾x4+5

   ⇔√x+4√x−11√x−1⩾x4+5⇔x+4x−11x−1⩾x4+5

   ⇔√x+4⩾x4+5⇔x+4⩾x4+5

   ⇔x4−√x+1⩾0⇔x4−x+1⩾0

   ⇔(√x2−1)2⩾0⇔(x2−1)2⩾0 (luôn đúng)

   Vậy x⩾0; x≠1

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