Trang chủ » Hỏi đáp » Môn Toán 2/( căn 2 + căn ( 2 + căn 3 ))+ 2/( căn 2 – căn ( 2 – căn 3 )) 13/11/2024 2/( căn 2 + căn ( 2 + căn 3 ))+ 2/( căn 2 – căn ( 2 – căn 3 ))
Giải đáp: $2\sqrt 2 $ Lời giải và giải thích chi tiết: $\begin{array}{l}\dfrac{2}{{\sqrt 2 + \sqrt {2 + \sqrt 3 } }} + \dfrac{2}{{\sqrt 2 – \sqrt {2 – \sqrt 3 } }}\\ = \dfrac{{2\sqrt 2 }}{{\sqrt 2 .\sqrt 2 + \sqrt 2 .\sqrt {2 + \sqrt 3 } }} + \dfrac{{2\sqrt 2 }}{{\sqrt 2 .\sqrt 2 – \sqrt 2 .\sqrt {2 – \sqrt 3 } }}\\ = \dfrac{{2\sqrt 2 }}{{2 + \sqrt {4 + 2\sqrt 3 } }} + \dfrac{{2\sqrt 2 }}{{2 – \sqrt {4 – 2\sqrt 3 } }}\\ = \dfrac{{2\sqrt 2 }}{{2 + \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }} + \dfrac{{2\sqrt 2 }}{{2 – \sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} }}\\ = \dfrac{{2\sqrt 2 }}{{2 + \sqrt 3 + 1}} + \dfrac{{2\sqrt 2 }}{{2 – \left( {\sqrt 3 – 1} \right)}}\\ = \dfrac{{2\sqrt 2 }}{{3 + \sqrt 3 }} + \dfrac{{2\sqrt 2 }}{{3 – \sqrt 3 }}\\ = \dfrac{{2\sqrt 2 .\left( {3 – \sqrt 3 } \right) + 2\sqrt 2 \left( {3 + \sqrt 3 } \right)}}{{\left( {3 – \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}}\\ = \dfrac{{6\sqrt 2 – 2\sqrt 6 + 6\sqrt 2 + 2\sqrt 6 }}{{{3^2} – 3}}\\ = \dfrac{{12\sqrt 2 }}{6}\\ = 2\sqrt 2 \end{array}$ Trả lời
\dfrac{2}{{\sqrt 2 + \sqrt {2 + \sqrt 3 } }} + \dfrac{2}{{\sqrt 2 – \sqrt {2 – \sqrt 3 } }}\\
= \dfrac{{2\sqrt 2 }}{{\sqrt 2 .\sqrt 2 + \sqrt 2 .\sqrt {2 + \sqrt 3 } }} + \dfrac{{2\sqrt 2 }}{{\sqrt 2 .\sqrt 2 – \sqrt 2 .\sqrt {2 – \sqrt 3 } }}\\
= \dfrac{{2\sqrt 2 }}{{2 + \sqrt {4 + 2\sqrt 3 } }} + \dfrac{{2\sqrt 2 }}{{2 – \sqrt {4 – 2\sqrt 3 } }}\\
= \dfrac{{2\sqrt 2 }}{{2 + \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }} + \dfrac{{2\sqrt 2 }}{{2 – \sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} }}\\
= \dfrac{{2\sqrt 2 }}{{2 + \sqrt 3 + 1}} + \dfrac{{2\sqrt 2 }}{{2 – \left( {\sqrt 3 – 1} \right)}}\\
= \dfrac{{2\sqrt 2 }}{{3 + \sqrt 3 }} + \dfrac{{2\sqrt 2 }}{{3 – \sqrt 3 }}\\
= \dfrac{{2\sqrt 2 .\left( {3 – \sqrt 3 } \right) + 2\sqrt 2 \left( {3 + \sqrt 3 } \right)}}{{\left( {3 – \sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}}\\
= \dfrac{{6\sqrt 2 – 2\sqrt 6 + 6\sqrt 2 + 2\sqrt 6 }}{{{3^2} – 3}}\\
= \dfrac{{12\sqrt 2 }}{6}\\
= 2\sqrt 2
\end{array}$