Đáp án:
\(b.\sqrt a \)
c. 2
Giải thích các bước giải:
\(\begin{array}{l}
a.\dfrac{{a - 1}}{{\sqrt a - 1}}.\dfrac{{\sqrt {{{\left( {\sqrt a - 1} \right)}^2}} }}{{{{\left( {a - 1} \right)}^2}}}\\
= \dfrac{{\left| {\sqrt a - 1} \right|}}{{\left( {\sqrt a - 1} \right)\left( {a - 1} \right)}} = \dfrac{{ - \left( {\sqrt a - 1} \right)}}{{\left( {\sqrt a - 1} \right)\left( {a - 1} \right)}}\left( {do:0 < a < 1} \right)\\
= - \dfrac{1}{{a - 1}}\\
c.2\left( {x + y} \right).\dfrac{1}{{\sqrt {{{\left( {x + y} \right)}^2}} }}\\
= \dfrac{{2\left( {x + y} \right)}}{{\left| {x + y} \right|}} = \dfrac{{2\left( {x + y} \right)}}{{x + y}}\left( {Do:x + y > 0} \right)\\
= 2\\
b.\dfrac{{a - b}}{{\sqrt {a + b} }}.\dfrac{{\sqrt {a\left( {a + b} \right)} }}{{\left| {a - b} \right|}} = \dfrac{{\left( {a - b} \right).\sqrt a }}{{\left( {a - b} \right)}}\left( {do:a > b > 0} \right)\\
= \sqrt a
\end{array}\)
