chi tiết đúng

a. Tập xác định: x ≥ 0 và x khác 1

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

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

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

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

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

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

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

b. Khi x = 4 thì: $D=\sqrt{4} .\left( 1-\sqrt{4}\right) =2.( 1-2) =-2$