Giải câu 1 đề 5 ôn thi toán lớp 9 lên 10

ĐỀ THI

Bài 1: (1,5 điểm)

Cho hai biểu thức:

$A=\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7 +2\sqrt{10}}}$

$B=\frac{15\sqrt{x}-11}{x +2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}$ với $x\geq 0, x\neq 1$

a. Rút gọn các biểu thức A và B

b. So sánh B với $\frac{2}{3}$

a. Rút gọn:

$A=\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7 +2\sqrt{10}}}$

$=\frac{\sqrt{(\sqrt{3}-\sqrt{2})^{2}}+\sqrt{(\sqrt{5}-\sqrt{3})^{2}}}{\sqrt{(\sqrt{5}+\sqrt{2})^{2}}}$

$=\frac{\left | \sqrt{3}-\sqrt{2} \right |+\left | \sqrt{5}-\sqrt{3} \right |}{\left | \sqrt{5}+\sqrt{2} \right |}$

$=\frac{\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$

$=\frac{(\sqrt{5}-\sqrt{2})^{2}}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}=\frac{7-2\sqrt{10}}{3}$

$B=\frac{15\sqrt{x}-11}{x +2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}$ với $x\geq 0, x\neq 1$

$=\frac{15\sqrt{x}-11}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}$

$=\frac{15\sqrt{x}-11-(3\sqrt{x}-2)(\sqrt{x}+3)-(2\sqrt{x}+3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+3)}$

$=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{(\sqrt{x}-1)(\sqrt{x}+3)}$

$=\frac{-5x + 7\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{(\sqrt{x}-1)(2-5\sqrt{x})}{(\sqrt{x}-1)(\sqrt{x}+3)}$

$=\frac{2-5\sqrt{x}}{\sqrt{x}+3}$

b. So sánh:

$B - \frac{2}{3}=\frac{2-5\sqrt{x}}{\sqrt{x}+3}-\frac{2}{3}=\frac{3(2-5\sqrt{x})-2(\sqrt{x}+3)}{3(\sqrt{x}+3)}=\frac{-17\sqrt{x}}{3(\sqrt{x}+3)}$

Do $\sqrt{x}\geq 0$ nên:

$\left\{\begin{matrix}3(\sqrt{x}+3) > 0& & \\ -17\sqrt{x}\leq 0& & \end{matrix}\right.$

$\Rightarrow \frac{-17\sqrt{x}}{3(\sqrt{x}+3)}\leq 0\Leftrightarrow B-\frac{2}{3}\leq 0\Leftrightarrow B\leq \frac{2}{3}$