P=A:B = (2x+1/x √x-1 - 1/ √x - 1): √x +3/x+ √x +1 a) rút gọn biểu thức b) tính giá trị B khi x = 16

Đáp án:

ĐKXĐ : $x ≥ 0 ; x \ne 1$

a. P = $\left(\dfrac{2x+1}{x\sqrt[]{x}-1} - \dfrac{1}{\sqrt[]{x}-1} \right )$ : $\dfrac{\sqrt[]{x}+3}{x+\sqrt[]{x}+1}$

⇔ P = $\left(\dfrac{2x+1}{(\sqrt[]{x})^{3}-1^{3}} - \dfrac{x+\sqrt[]{x}+1}{(\sqrt[]{x}-1)(x+\sqrt[]{x}+1)} \right )$ . $\dfrac{x+\sqrt[]{x}+1}{\sqrt[]{x}+3}$

⇔ P = $\left(\dfrac{2x+1}{(\sqrt[]{x}-1)(x+\sqrt[]{x}+1)} - \dfrac{x+\sqrt[]{x}+1}{(\sqrt[]{x}-1)(x+\sqrt[]{x}+1)} \right )$ . $\dfrac{x+\sqrt[]{x}+1}{\sqrt[]{x}+3}$

⇔ P = $\dfrac{x-\sqrt[]{x}}{(\sqrt[]{x}-1)(x+\sqrt[]{x}+1)} . \dfrac{x+\sqrt[]{x}+1}{\sqrt[]{x}+3}$

⇔ P = $\dfrac{\sqrt[]{x}(\sqrt[]{x}-1)}{(\sqrt[]{x}-1)(x+\sqrt[]{x}+1)} . \dfrac{x+\sqrt[]{x}+1}{\sqrt[]{x}+3}$

⇔ P = $\dfrac{\sqrt[]{x}}{\sqrt[]{x}+3}$

b. $x = 16$ thay vào $B :$

$B = \dfrac{\sqrt[]{16}+3}{16+\sqrt[]{16}+1}$

⇔ $B = \dfrac{4+3}{16+4+1}$

⇔ $B = \dfrac{7}{21}$

⇔ $B = \dfrac{1}{3}$