a) Vì \(x ≥ 0\) nên \( x = {\left( {\sqrt x } \right)^2}\)
Ta có:
\( \displaystyle\eqalign{
& \sqrt {{{x - 2\sqrt x + 1} \over {x + 2\sqrt x + 1}}} \cr
& = \sqrt {{{{{\left( {\sqrt x } \right)}^2} - 2\sqrt x + 1} \over {{{\left( {\sqrt x } \right)}^2} + 2\sqrt x + 1}}} \cr
& = \sqrt {{{{{\left( {\sqrt x - 1} \right)}^2}} \over {{{\left( {\sqrt x + 1} \right)}^2}}}} \cr} \)
\( \displaystyle \displaystyle= {{\sqrt {{{\left( {\sqrt x - 1} \right)}^2}} } \over {\sqrt {{{\left( {\sqrt x + 1} \right)}^2}} }}\)
\( = \dfrac{{\left| {\sqrt x - 1} \right|}}{{\left| {\sqrt x + 1} \right|}} = \dfrac{{\left| {\sqrt x - 1} \right|}}{{\sqrt x + 1}}\)
+) Nếu \( \displaystyle\sqrt x - 1 \ge 0 \Leftrightarrow x \ge 1\) thì \( \displaystyle\left| {\sqrt x - 1} \right| = \sqrt x - 1\)
Ta có: \( \displaystyle{{\left| {\sqrt x - 1} \right|} \over {\sqrt x + 1}} = {{\sqrt x - 1} \over {\sqrt x + 1}}\) (với \(x ≥ 1)\)
+) Nếu \( \displaystyle\sqrt x - 1 < 0 \Leftrightarrow x < 1\) thì \( \displaystyle\left| {\sqrt x - 1} \right| = 1 - \sqrt x \)
Ta có:
\( \displaystyle{{\left| {\sqrt x - 1} \right|} \over {\sqrt x + 1}} = {{1 - \sqrt x } \over {\sqrt x + 1}}\) (với \(0 ≤ x < 1\))
b) Vì \(y ≥ 0\) nên \( y = {\left( {\sqrt y } \right)^2}\)
Ta có:
\( \displaystyle\eqalign{
& {{x - 1} \over {\sqrt y - 1}}\sqrt {{{{{ {y - 2\sqrt y + 1} }}} \over {{{(x - 1)}^4}}}} \cr
& = {{x - 1} \over {\sqrt y - 1}}{{\sqrt {{{\left( \sqrt y - 1 \right)}^2}} } \over {\sqrt {{{(x - 1)}^4}} }} \cr} \)
\( \displaystyle\eqalign{
& = {{x - 1} \over {\sqrt y - 1}}.{{\left| \sqrt y-1 \right|} \over {{{(x - 1)}^2}}} \cr} \)
+) Nếu \(y>1\)
Ta có \( \displaystyle\left| \sqrt y-1 \right|=\sqrt y-1\) nên ta có kết quả là \( \displaystyle{{x - 1} \over {\sqrt y - 1}}.{{ \sqrt y-1 } \over {{{(x - 1)}^2}}} =\dfrac {1}{x-1}\)
+) Nếu \(0 \le y < 1\)
Ta có \(\left| {\sqrt y - 1} \right| = -( \sqrt y -1)\) nên ta có kết quả là
\(\dfrac{{x - 1}}{{\sqrt y - 1}}.\dfrac{{-(\sqrt y -1)}}{{{{(x - 1)}^2}}} = \dfrac{{ - 1}}{{x - 1}}\)
