$\frac{1}{ √5 - √6}$ . √ $\frac{6√5 - 5√6}{6√5 + 5√6}$
1 câu trả lời
Đáp án:
\[ - 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{1}{{\sqrt 5 - \sqrt 6 }}.\sqrt {\dfrac{{6\sqrt 5 - 5\sqrt 6 }}{{6\sqrt 5 + 5\sqrt 6 }}} \\
= \dfrac{1}{{\sqrt 5 - \sqrt 6 }}.\sqrt {\dfrac{{{{\sqrt 6 }^2}.\sqrt 5 - {{\sqrt 5 }^2}.\sqrt 6 }}{{{{\sqrt 6 }^2}.\sqrt 5 + {{\sqrt 5 }^2}.\sqrt 6 }}} \\
= \dfrac{1}{{\sqrt 5 - \sqrt 6 }}.\sqrt {\dfrac{{\sqrt 6 .\sqrt 5 .\left( {\sqrt 6 - \sqrt 5 } \right)}}{{\sqrt 6 .\sqrt 5 .\left( {\sqrt 6 + \sqrt 5 } \right)}}} \\
= \dfrac{1}{{\sqrt 5 - \sqrt 6 }}.\sqrt {\dfrac{{\sqrt 6 - \sqrt 5 }}{{\sqrt 6 + \sqrt 5 }}} \\
= \dfrac{1}{{\sqrt 5 - \sqrt 6 }}.\sqrt {\dfrac{{{{\left( {\sqrt 6 - \sqrt 5 } \right)}^2}}}{{\left( {\sqrt 6 + \sqrt 5 } \right).\left( {\sqrt 6 - \sqrt 5 } \right)}}} \\
= \dfrac{1}{{\sqrt 5 - \sqrt 6 }}.\sqrt {\dfrac{{{{\left( {\sqrt 6 - \sqrt 5 } \right)}^2}}}{{{{\sqrt 6 }^2} - {{\sqrt 5 }^2}}}} \\
= \dfrac{1}{{\sqrt 5 - \sqrt 6 }}.\sqrt {\dfrac{{{{\left( {\sqrt 6 - \sqrt 5 } \right)}^2}}}{1}} \\
= \dfrac{1}{{\sqrt 5 - \sqrt 6 }}.\left| {\sqrt 6 - \sqrt 5 } \right|\\
= \dfrac{1}{{\sqrt 5 - \sqrt 6 }}.\left( {\sqrt 6 - \sqrt 5 } \right)\\
= - 1
\end{array}\)