Đáp án:
a) $A=\dfrac{x-1}{\sqrt{x}}$
b) $x=3+2\sqrt{2}$ thì $A=2$
c) $A=\dfrac{-3+\sqrt{3}}{2}$ khi $x=4-2\sqrt{3}$
Giải thích các bước giải:
a)
$A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right) (x>0, x\ne 1)\\A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right)\\A=\dfrac{x-1}{\sqrt{x}(\sqrt{x}-1)}:\dfrac{\sqrt{x}-1+2}{(\sqrt{x}-1)(\sqrt{x}+1)}\\A=\dfrac{x-1}{\sqrt{x}(\sqrt{x}-1)}.\dfrac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}+1}\\A=\dfrac{x-1}{\sqrt{x}}$
b)
$A=\dfrac{x-1}{\sqrt{x}} (x>0)\\A=2\to \dfrac{x-1}{\sqrt{x}}=2\\\to x-1=2\sqrt{x}\\\to x-2\sqrt{x}-1=0$
Đặt $t=\sqrt{x} (t\ge0)\\\to t^2-2t-1=0\\\Delta'=(-1)^2-(-1)=2\\\to t_1=1+\sqrt{2} \text{ (thoả mãn)}\\t_2=1-\sqrt{2} \text{ (loại)}\\\to \sqrt{x}=1+\sqrt{2}\\\to x=(1+\sqrt{2})^2=3+2\sqrt{2}$
Vậy $x=3+2\sqrt{2}$
c)
$A=\dfrac{x-1}{\sqrt{x}}$
Khi $x=4-2\sqrt{3}=(\sqrt{3}-1)^2 \to A=\dfrac{4-2\sqrt{3}-1}{\sqrt{4-2\sqrt{3}}}\\A=\dfrac{3-2\sqrt{3}}{\sqrt{(\sqrt{3}-1)^2}}\\A=\dfrac{3-2\sqrt{3}}{|\sqrt{3}-1|}\\A=\dfrac{3-2\sqrt{3}}{\sqrt{3}-1}\\A=\dfrac{(3-2\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\A=\dfrac{-3+\sqrt{3}}{2}$
Vậy $A=\dfrac{-3+\sqrt{3}}{2}$ khi $x=4-2\sqrt{3}$
