Tính giá trị biểu thức :
A = $\frac{3+\sqrt[]{5} }{\sqrt[]{10} + \sqrt[]{3+\sqrt[]{5}}}$ – $\frac{3-\sqrt[]{5}}{\sqrt[]{10} – \sqrt[]{3-\sqrt[]{5}}}$
Tính giá trị biểu thức :
A = $\frac{3+\sqrt[]{5} }{\sqrt[]{10} + \sqrt[]{3+\sqrt[]{5}}}$ – $\frac{3-\sqrt[]{5}}{\sqrt[]{10} – \sqrt[]{3-\sqrt[]{5}}}$
Đáp án:
$A = \dfrac{{15\sqrt {10} – 5\sqrt 2 }}{{22}}$
Giải thích các bước giải:
$\begin{array}{l}
A = \dfrac{{3 + \sqrt 5 }}{{\sqrt {10} + \sqrt {3 + \sqrt 5 } }} – \dfrac{{3 – \sqrt 5 }}{{\sqrt {10} – \sqrt {3 – \sqrt 5 } }}\\
= \dfrac{{\sqrt 2 \left( {3 + \sqrt 5 } \right)}}{{2\sqrt 5 + \sqrt {6 + 2\sqrt 5 } }} – \dfrac{{\sqrt 2 \left( {3 – \sqrt 5 } \right)}}{{2\sqrt 5 – \sqrt {6 – 2\sqrt 5 } }}\\
= \dfrac{{\sqrt 2 \left( {3 + \sqrt 5 } \right)}}{{2\sqrt 5 + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }} – \dfrac{{\sqrt 2 \left( {3 – \sqrt 5 } \right)}}{{2\sqrt 5 – \sqrt {{{\left( {\sqrt 5 – 1} \right)}^2}} }}\\
= \dfrac{{3\sqrt 2 + \sqrt {10} }}{{2\sqrt 5 + \sqrt 5 + 1}} – \dfrac{{3\sqrt 2 – \sqrt {10} }}{{2\sqrt 5 – \left( {\sqrt 5 – 1} \right)}}\\
= \dfrac{{3\sqrt 2 + \sqrt {10} }}{{3\sqrt 5 + 1}} – \dfrac{{3\sqrt 2 – \sqrt {10} }}{{1 – \sqrt 5 }}\\
= \dfrac{{\left( {3\sqrt 2 + \sqrt {10} } \right)\left( {3\sqrt 5 – 1} \right)}}{{\left( {3\sqrt 5 + 1} \right)\left( {3\sqrt 5 – 1} \right)}} – \dfrac{{\left( {3\sqrt 2 – \sqrt {10} } \right)\left( {1 + \sqrt 5 } \right)}}{{\left( {1 – \sqrt 5 } \right)\left( {1 + \sqrt 5 } \right)}}\\
= \dfrac{{9\sqrt {10} – 3\sqrt 2 + 15\sqrt 2 – \sqrt {10} }}{{{{\left( {3\sqrt 5 } \right)}^2} – {1^2}}} – \dfrac{{3\sqrt 2 + 3\sqrt {10} – \sqrt {10} – 5\sqrt 2 }}{{{1^2} – {{\left( {\sqrt 5 } \right)}^2}}}\\
= \dfrac{{12\sqrt 2 + 8\sqrt {10} }}{{44}} – \dfrac{{2\sqrt {10} – 2\sqrt 2 }}{{ – 4}}\\
= \dfrac{{3\sqrt 2 + 2\sqrt {10} }}{{11}} + \dfrac{{\sqrt {10} – \sqrt 2 }}{2}\\
= \dfrac{{2\left( {3\sqrt 2 + 2\sqrt {10} } \right) + 11\left( {\sqrt {10} – \sqrt 2 } \right)}}{{22}}\\
= \dfrac{{6\sqrt 2 + 4\sqrt {10} + 11\sqrt {10} – 11\sqrt 2 }}{{22}}\\
= \dfrac{{15\sqrt {10} – 5\sqrt 2 }}{{22}}
\end{array}$
Vậy $A = \dfrac{{15\sqrt {10} – 5\sqrt 2 }}{{22}}$
Đáp án:
\(\dfrac{{ – \sqrt {10} + 5\sqrt 2 }}{{4 + \sqrt 5 }}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{{3 + \sqrt 5 }}{{\sqrt {10} + \sqrt {3 + \sqrt 5 } }} – \dfrac{{3 – \sqrt 5 }}{{\sqrt {10} – \sqrt {3 – \sqrt 5 } }}\\
= \dfrac{{3\sqrt 2 + \sqrt {10} }}{{2\sqrt 5 + \sqrt {6 + 2\sqrt 5 } }} – \dfrac{{3\sqrt 2 – \sqrt {10} }}{{2\sqrt 5 – \sqrt {6 – 2\sqrt 5 } }}\\
= \dfrac{{3\sqrt 2 + \sqrt {10} }}{{2\sqrt 5 + \sqrt {5 + 2\sqrt 5 .1 + 1} }} – \dfrac{{3\sqrt 2 – \sqrt {10} }}{{2\sqrt 5 – \sqrt {5 – 2\sqrt 5 .1 + 1} }}\\
= \dfrac{{3\sqrt 2 + \sqrt {10} }}{{2\sqrt 5 + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }} – \dfrac{{3\sqrt 2 – \sqrt {10} }}{{2\sqrt 5 – \sqrt {{{\left( {\sqrt 5 – 1} \right)}^2}} }}\\
= \dfrac{{3\sqrt 2 + \sqrt {10} }}{{2\sqrt 5 + \sqrt 5 + 1}} – \dfrac{{3\sqrt 2 – \sqrt {10} }}{{2\sqrt 5 – \sqrt 5 + 1}}\\
= \dfrac{{3\sqrt 2 + \sqrt {10} }}{{3\sqrt 5 + 1}} – \dfrac{{3\sqrt 2 – \sqrt {10} }}{{\sqrt 5 + 1}}\\
= \dfrac{{\left( {3\sqrt 2 + \sqrt {10} } \right)\left( {\sqrt 5 + 1} \right) – \left( {3\sqrt 2 – \sqrt {10} } \right)\left( {3\sqrt 5 + 1} \right)}}{{\left( {3\sqrt 5 + 1} \right)\left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{3\sqrt {10} + 3\sqrt 2 + 5\sqrt 2 + \sqrt {10} – 9\sqrt {10} – 3\sqrt 2 + 15\sqrt 2 + \sqrt {10} }}{{15 + 3\sqrt 5 + \sqrt 5 + 1}}\\
= \dfrac{{ – 4\sqrt {10} + 20\sqrt 2 }}{{16 + 4\sqrt 5 }}\\
= \dfrac{{ – \sqrt {10} + 5\sqrt 2 }}{{4 + \sqrt 5 }}
\end{array}\)