Đáp án:
a) $A=\dfrac{3}{2\sqrt{x}}$
b) $A=\dfrac{3\sqrt{5}+3}{8}$
Giải thích các bước giải:
Với $x>0,x\ne 1$
a) $A=\left( \dfrac{1}{1-\sqrt{x}}+\dfrac{1}{1+\sqrt{x}} \right):\left( \dfrac{1}{1-\sqrt{x}}-\dfrac{1}{1+\sqrt{x}} \right)+\dfrac{1}{2\sqrt{x}}$
$A=\dfrac{1+\sqrt{x}+1-\sqrt{x}}{\left( 1-\sqrt{x} \right)\left( 1+\sqrt{x} \right)}:\dfrac{1+\sqrt{x}-1+\sqrt{x}}{\left( 1-\sqrt{x} \right)\left( 1 \right)+\sqrt{x}}+\dfrac{1}{2\sqrt{x}}$
$A=\dfrac{2}{\left( 1-\sqrt{x} \right)\left( 1+\sqrt{x} \right)}\cdot\dfrac{\left( 1-\sqrt{x} \right)\left( 1+\sqrt{x} \right)}{2\sqrt{x}}+\dfrac{1}{2\sqrt{x}}$
$A=\dfrac{1}{\sqrt{x}}+\dfrac{1}{2\sqrt{x}}$
$A=\dfrac{3}{2\sqrt{x}}$
b) Với $x=6-2\sqrt{5}$
$A=\dfrac{3}{2\sqrt{6-2\sqrt{5}}}$
$A=\dfrac{3}{2\sqrt{{{\left( \sqrt{5}-1 \right)}^{2}}}}$
$A=\dfrac{3}{2\left| \sqrt{5}-1 \right|}$
$A=\dfrac{3}{2\left( \sqrt{5}-1 \right)}$
$A=\dfrac{3\left( \sqrt{5}+1 \right)}{2\left( \sqrt{5}-1 \right)\left( \sqrt{5}+1 \right)}$
$A=\dfrac{3\sqrt{5}+3}{8}$
