Упрощение алгебраических выражений. Задача 61

ОТВЕТ: \(\dfrac{{{b^2}-1}}{{\sqrt b }},\)  если     \(b \in \left( {0;\,1} \right);\)  \(\dfrac{{{b^2} + 3}}{{\sqrt b }},\)  если  \(b \in \left( {1; + \infty } \right)\)

\(\dfrac{{\dfrac{{\left| {b-1} \right|}}{b} + b\left| {b-1} \right| + 2-\dfrac{2}{b}}}{{\sqrt {b-2 + \dfrac{1}{b}} }} = \dfrac{{\left| {b-1} \right| + {b^2}\left| {b-1} \right| + 2b-2}}{{b\sqrt {\dfrac{{{b^2}-2b + 1}}{b}} }} = \)

\( = \dfrac{{\left| {b-1} \right| + {b^2}\left| {b-1} \right| + 2b-2}}{{{{\left( {\sqrt b } \right)}^2} \cdot \dfrac{{\sqrt {{{\left( {b-1} \right)}^2}} }}{{\sqrt b }}}} = \dfrac{{\left| {b-1} \right| + {b^2}\left| {b-1} \right| + 2b-2}}{{\sqrt b \left| {b-1} \right|}}.\)

Если \(b\, \in \,\left( {0;1} \right)\), то

\(\dfrac{{\left| {b-1} \right| + {b^2}\left| {b-1} \right| + 2b-2}}{{\sqrt b \left| {b-1} \right|}} = \dfrac{{-\left( {b-1} \right)-{b^2}\left( {b-1} \right) + 2\left( {b-1} \right)}}{{-\sqrt b \left( {b-1} \right)}} = \dfrac{{\left( {b-1} \right)\left( {-1-{b^2} + 2} \right)}}{{-\sqrt b \left( {b-1} \right)}} = \dfrac{{{b^2}-1}}{{\sqrt b }}.\)

Если \(b\, \in \,\left( {1;\infty } \right)\), то

\(\dfrac{{\left| {b-1} \right| + {b^2}\left| {b-1} \right| + 2b-2}}{{\sqrt b \left| {b-1} \right|}} = \dfrac{{b-1 + {b^2}\left( {b-1} \right) + 2\left( {b-1} \right)}}{{\sqrt b \left( {b-1} \right)}} = \dfrac{{\left( {b-1} \right)\left( {1 + {b^2} + 2} \right)}}{{\sqrt b \left( {b-1} \right)}} = \dfrac{{{b^2} + 3}}{{\sqrt b }}.\)

Ответ:  \(\dfrac{{{b^2}-1}}{{\sqrt b }},\)  если   \(b \in \left( {0;\,1} \right);\)   \(\dfrac{{{b^2} + 3}}{{\sqrt b }},\)  если  \(b \in \left( {1; + \infty } \right).\)