Rút gọn biểu thức \(A = \sqrt {1 + \dfrac{1}{{{a^2}}} + \dfrac{1}{{{{\left( {a + 1} \right)}^2}}}} \) với \(\left( {a > 0} \right)\)

\(\begin{array}{l}A = \sqrt {1 + \dfrac{1}{{{a^2}}} + \dfrac{1}{{{{\left( {a + 1} \right)}^2}}}} \\ \Rightarrow {A^2} = {\left( {\sqrt {1 + \dfrac{1}{{{a^2}}} + \dfrac{1}{{{{\left( {a + 1} \right)}^2}}}} } \right)^2}\\ = 1 + \dfrac{1}{{{a^2}}} + \dfrac{1}{{{{\left( {a + 1} \right)}^2}}}\\ = \dfrac{{{a^2}{{\left( {a + 1} \right)}^2} + {{\left( {a + 1} \right)}^2} + {a^2}}}{{{a^2}{{\left( {a + 1} \right)}^2}}}\\ = \dfrac{{{a^2}\left( {{a^2} + 2a + 1 + 1} \right) + {{\left( {a + 1} \right)}^2}}}{{{a^2}{{\left( {a + 1} \right)}^2}}}\\ = \dfrac{{{a^4} + 2{a^2}\left( {a + 1} \right) + {{\left( {a + 1} \right)}^2}}}{{{a^2}{{\left( {a + 1} \right)}^2}}}\\ = \dfrac{{{{\left( {{a^2} + a + 1} \right)}^2}}}{{{a^2}{{\left( {a + 1} \right)}^2}}} = {\left[ {\dfrac{{{a^2} + a + 1}}{{a\left( {a + 1} \right)}}} \right]^2}.\end{array}\)

Do \(a > 0\) nên \(A > 0\) và \(A = \dfrac{{{a^2} + a + 1}}{{a\left( {a + 1} \right)}}.\)