Đáp án:
Bài 8: Rút gọn
$A=\dfrac{2+\sqrt{3}}{\sqrt{7-4\sqrt{3}}}-\dfrac{2-\sqrt{3}}{\sqrt{7+4\sqrt{3}}}$
$A=\dfrac{2+\sqrt{3}}{\sqrt{{{\left( 2-\sqrt{3} \right)}^{2}}}}-\dfrac{2-\sqrt{3}}{\sqrt{{{\left( 2+\sqrt{3} \right)}^{2}}}}$
$A=\dfrac{2+\sqrt{3}}{\left| 2-\sqrt{3} \right|}-\dfrac{2-\sqrt{3}}{\left| 2+\sqrt{3} \right|}$
$A=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}-\dfrac{2-\sqrt{3}}{2+\sqrt{3}}$
$A=\dfrac{{{\left( 2+\sqrt{3} \right)}^{2}}-{{\left( 2-\sqrt{3} \right)}^{2}}}{\left( 2-\sqrt{3} \right)\left( 2+\sqrt{3} \right)}$
$A=\dfrac{7+4\sqrt{3}-\left( 7-4\sqrt{3} \right)}{1}$
$A=8\sqrt{3}$
Bài 9:
$A=\dfrac{\sqrt{x}}{\sqrt{x}+3}$ và $B=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}$ với $x\ge 0,x\ne 9$
a) Tính giá trị biểu thức $A$ khi $x=16$
Với $x=16\Rightarrow A=\dfrac{\sqrt{16}}{\sqrt{16}+3}=\dfrac{4}{4+3}=\dfrac{4}{7}$
b) Chứng minh $A+B=\dfrac{3}{\sqrt{x}+3}$
$B=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}$
$B=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{\left( \sqrt{x}-3 \right)\left( \sqrt{x}+3 \right)}$
$B=\dfrac{2\sqrt{x}\left( \sqrt{x}+3 \right)-\left( 3x+9 \right)}{\left( \sqrt{x}-3 \right)\left( \sqrt{x}+3 \right)}$
$B=\dfrac{2x+6\sqrt{x}-3x-9}{\left( \sqrt{x}-3 \right)\left( \sqrt{x}+3 \right)}$
$B=\dfrac{-x+6\sqrt{x}-9}{\left( \sqrt{x}-3 \right)\left( \sqrt{x}+3 \right)}$
$B=\dfrac{-{{\left( \sqrt{x}-3 \right)}^{2}}}{\left( \sqrt{x}-3 \right)\left( \sqrt{x}+3 \right)}$
$B=\dfrac{-\left( \sqrt{x}-3 \right)}{\sqrt{x}+3}$
$VT=A+B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{-\left( \sqrt{x}-3 \right)}{\sqrt{x}+3}=\dfrac{\sqrt{x}-\sqrt{x}+3}{\sqrt{x}+3}=\dfrac{3}{\sqrt{x}+3}=VP$
