Rút gọn biểu thức chứa căn thức bậc hai SVIP

Hướng dẫn giải:

$B=\Big[\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}\Big].\dfrac{1}{\sqrt{x}+1}$

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

$=\dfrac{1}{\sqrt{x}}$.

Hướng dẫn giải:

$P=\Big[\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\Big] \, : \, \dfrac{1-\sqrt{x}}{2-\sqrt{x}}$

$=\Big[\dfrac{x-\sqrt{x}+2}{(\sqrt{x}+1)(\sqrt{x}-2)}-\dfrac{x}{\sqrt{x}(\sqrt{x}-2)}\Big] \, : \, \dfrac{1-\sqrt{x}}{2-\sqrt{x}}$

$=\dfrac{x\sqrt{x}-x+2\sqrt{x}-x(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}-2)} \, : \, \dfrac{1-\sqrt{x}}{2-\sqrt{x}}$

$=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}-2)}.\dfrac{2-\sqrt{x}}{1-\sqrt{x}}$

$=\dfrac{2\sqrt{x}(1-\sqrt{x})}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}-2)}.\dfrac{2-\sqrt{x}}{1-\sqrt{x}}$

$=\dfrac{-2}{\sqrt{x}+1}.$

Hướng dẫn giải:

$A=\dfrac{2a^2+4}{1-a^2}-\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}$

$=\dfrac{2a^2+4}{1-a^2}-\dfrac{1+\sqrt{a}+1-\sqrt{a}}{1-a}$

$=\dfrac{2a^2+4}{1-a^2}-\dfrac{2}{1-a}$

$=\dfrac{2a^2+4-2(1+a)}{1-a^2}$

$=\dfrac{2a^2-2a+2}{1-a^2}$.

Hướng dẫn giải:

$P=\dfrac{x+2}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}$

$=\dfrac{x+2-\left(\sqrt{x}+2\right)+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}$

$=\dfrac{x}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+2}$

Hướng dẫn giải:

$P=\Big(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\Big)\Big(\sqrt{x}-\dfrac{1}{\sqrt{x}}\Big)$

$=\dfrac{\sqrt{x}(\sqrt{x}-1)+\sqrt{x}(\sqrt{x}+1)}{x-1}.\Big(\dfrac{x-1}{\sqrt{x}}\Big)$

$=\dfrac{2x}{x-1}.\dfrac{x-1}{\sqrt{x}}=2\sqrt{x}$.

Hướng dẫn giải:

$B=\Big(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{\sqrt{x}}\Big) \, : \, \dfrac{\sqrt{x}}{\sqrt{x}+1}$

$=\dfrac{x-(x-1)}{\sqrt{x}(\sqrt{x}+1)} \, : \, \dfrac{\sqrt{x}}{\sqrt{x}+1}$

$=\dfrac{x-(x-1)}{\sqrt{x}(\sqrt{x}+1)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{1}{x}$.

Hướng dẫn giải:

$B=\Big(\dfrac{x}{\sqrt{x}+3}-\dfrac{x+1}{\sqrt{x}-3}+\dfrac{6x+\sqrt{x}}{x-9}\Big) \, : \, \Big(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-1\Big)$

$=\Big(\dfrac{x}{\sqrt{x}+3}-\dfrac{x+1}{\sqrt{x}-3}+\dfrac{6x+\sqrt{x}}{x-9}\Big) \, : \, \Big(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-1\Big)$

$=\dfrac{x(\sqrt{x}-3)-(x+1)(\sqrt{x}+3)+6x+\sqrt{x}}{x-9} \, : \, \dfrac{\sqrt{x}-3-\sqrt{x}-3}{\sqrt{x}+3}$

$=\dfrac{-3}{x-9} \, : \, \dfrac{-6}{\sqrt{x}+3}$

$=\dfrac{-3}{x-9}.\dfrac{\sqrt{x}+3}{-6}$

$=\dfrac{1}{2(\sqrt{x}-3)}.$

Hướng dẫn giải:

$P=\dfrac{1}{x^2-\sqrt{x}} \, : \, \dfrac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}$

$=\dfrac{1}{\sqrt{x}(x\sqrt{x}-1)} \, : \, \dfrac{\sqrt{x}+1}{\sqrt{x}(x+\sqrt{x}+1)}$

$=\dfrac{1}{\sqrt{x}(x\sqrt{x}-1)}.\dfrac{\sqrt{x}(x+\sqrt{x}+1)}{\sqrt{x}+1}$

$=\dfrac{1}{x-1}$.

Hướng dẫn giải:

$V=\Big(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\Big).\dfrac{\sqrt{x}+2}{\sqrt{x}}$

$=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{x-4}.\dfrac{\sqrt{x}+2}{\sqrt{x}}$

$=\dfrac{2\sqrt{x}}{x-4}.\dfrac{\sqrt{x}+2}{\sqrt{x}}$

$=\dfrac{2}{\sqrt{x}-2}$.

Hướng dẫn giải:

$P=\Big(1+\dfrac{1}{\sqrt{x}}\Big)\Big(\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\Big)$

$=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\dfrac{\sqrt{x}-1+\sqrt{x}+1-2}{x-1}$

$=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\dfrac{2\left(\sqrt{x}-1\right)}{x-1}$

$=\dfrac{2}{\sqrt{x}}$.