Đáp án: 6) F=-1 Giải thích các bước giải: \(\begin{array}{l}5)E = \dfrac{1}{{\sqrt {3 – 2\sqrt 3 .1 + 1} }} – \dfrac{1}{{\sqrt {7 – 4\sqrt 3 } }} + \dfrac{3}{{\sqrt {9 – 2.3.\sqrt 5 + 5} }}\\ = \dfrac{1}{{\sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} }} – \dfrac{1}{{\sqrt {4 – 2.2.\sqrt 3 + 3} }} + \dfrac{3}{{\sqrt {{{\left( {3 – \sqrt 5 } \right)}^2}} }}\\ = \dfrac{1}{{\sqrt 3 – 1}} – \dfrac{1}{{\sqrt {{{\left( {2 – \sqrt 3 } \right)}^2}} }} + \dfrac{3}{{3 – \sqrt 5 }}\\ = \dfrac{{\sqrt 3 + 1}}{{3 – 1}} – \dfrac{1}{{2 – \sqrt 3 }} + \dfrac{{3\left( {3 + \sqrt 5 } \right)}}{{9 – 5}}\\ = \dfrac{{\sqrt 3 + 1}}{2} – \dfrac{{2 + \sqrt 3 }}{{4 – 3}} + \dfrac{{9 + 3\sqrt 5 }}{4}\\ = \dfrac{{2\sqrt 3 + 2 – 8 – 4\sqrt 3 + 9 + 3\sqrt 5 }}{4}\\ = \dfrac{{3\sqrt 5 – 2\sqrt 3 + 3}}{4}\\6)F = \dfrac{{\sqrt 2 + \sqrt 3 }}{{2 – 3}}.\sqrt {\dfrac{{\sqrt 6 \left( {\sqrt 3 – \sqrt 2 } \right)}}{{\sqrt 6 \left( {\sqrt 3 + \sqrt 2 } \right)}}} \\ = \dfrac{{\sqrt 2 + \sqrt 3 }}{{ – 1}}.\sqrt {\dfrac{{\sqrt 3 – \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \\ = – \sqrt {\sqrt 3 + \sqrt 2 } .\sqrt {\sqrt 3 – \sqrt 2 } \\ = – \sqrt {3 – 2} = – 1\\7)G = \dfrac{{\sqrt {10 – 2.\sqrt {10} .\sqrt 5 + 5} + \sqrt {13 + 4\sqrt {10} – \sqrt {10 – 2\sqrt {10} .1 + 1} } }}{{2\sqrt {2 + 2\sqrt 2 .1 + 1} + \sqrt {9 – 4\sqrt 2 + \sqrt {8 + 2.2\sqrt 2 .2 + 4} } }}\\ = \dfrac{{\sqrt {{{\left( {\sqrt {10} – \sqrt 5 } \right)}^2}} + \sqrt {13 + 4\sqrt {10} – \sqrt {{{\left( {\sqrt {10} – 1} \right)}^2}} } }}{{2\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + \sqrt {9 – 4\sqrt 2 + \sqrt {{{\left( {2\sqrt 2 + 2} \right)}^2}} } }}\\ = \dfrac{{\sqrt {10} – \sqrt 5 + \sqrt {13 + 4\sqrt {10} – \sqrt {10} + 1} }}{{2\left( {\sqrt 2 + 1} \right) + \sqrt {9 – 4\sqrt 2 + 2\sqrt 2 + 2} }}\\ = \dfrac{{\sqrt {10} – \sqrt 5 + \sqrt {14 + 3\sqrt {10} } }}{{2\sqrt 2 + 2 + \sqrt {11 – 2\sqrt 2 } }}\end{array}\) Reply
Đáp án:
6) F=-1
Giải thích các bước giải:
\(\begin{array}{l}
5)E = \dfrac{1}{{\sqrt {3 – 2\sqrt 3 .1 + 1} }} – \dfrac{1}{{\sqrt {7 – 4\sqrt 3 } }} + \dfrac{3}{{\sqrt {9 – 2.3.\sqrt 5 + 5} }}\\
= \dfrac{1}{{\sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} }} – \dfrac{1}{{\sqrt {4 – 2.2.\sqrt 3 + 3} }} + \dfrac{3}{{\sqrt {{{\left( {3 – \sqrt 5 } \right)}^2}} }}\\
= \dfrac{1}{{\sqrt 3 – 1}} – \dfrac{1}{{\sqrt {{{\left( {2 – \sqrt 3 } \right)}^2}} }} + \dfrac{3}{{3 – \sqrt 5 }}\\
= \dfrac{{\sqrt 3 + 1}}{{3 – 1}} – \dfrac{1}{{2 – \sqrt 3 }} + \dfrac{{3\left( {3 + \sqrt 5 } \right)}}{{9 – 5}}\\
= \dfrac{{\sqrt 3 + 1}}{2} – \dfrac{{2 + \sqrt 3 }}{{4 – 3}} + \dfrac{{9 + 3\sqrt 5 }}{4}\\
= \dfrac{{2\sqrt 3 + 2 – 8 – 4\sqrt 3 + 9 + 3\sqrt 5 }}{4}\\
= \dfrac{{3\sqrt 5 – 2\sqrt 3 + 3}}{4}\\
6)F = \dfrac{{\sqrt 2 + \sqrt 3 }}{{2 – 3}}.\sqrt {\dfrac{{\sqrt 6 \left( {\sqrt 3 – \sqrt 2 } \right)}}{{\sqrt 6 \left( {\sqrt 3 + \sqrt 2 } \right)}}} \\
= \dfrac{{\sqrt 2 + \sqrt 3 }}{{ – 1}}.\sqrt {\dfrac{{\sqrt 3 – \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \\
= – \sqrt {\sqrt 3 + \sqrt 2 } .\sqrt {\sqrt 3 – \sqrt 2 } \\
= – \sqrt {3 – 2} = – 1\\
7)G = \dfrac{{\sqrt {10 – 2.\sqrt {10} .\sqrt 5 + 5} + \sqrt {13 + 4\sqrt {10} – \sqrt {10 – 2\sqrt {10} .1 + 1} } }}{{2\sqrt {2 + 2\sqrt 2 .1 + 1} + \sqrt {9 – 4\sqrt 2 + \sqrt {8 + 2.2\sqrt 2 .2 + 4} } }}\\
= \dfrac{{\sqrt {{{\left( {\sqrt {10} – \sqrt 5 } \right)}^2}} + \sqrt {13 + 4\sqrt {10} – \sqrt {{{\left( {\sqrt {10} – 1} \right)}^2}} } }}{{2\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + \sqrt {9 – 4\sqrt 2 + \sqrt {{{\left( {2\sqrt 2 + 2} \right)}^2}} } }}\\
= \dfrac{{\sqrt {10} – \sqrt 5 + \sqrt {13 + 4\sqrt {10} – \sqrt {10} + 1} }}{{2\left( {\sqrt 2 + 1} \right) + \sqrt {9 – 4\sqrt 2 + 2\sqrt 2 + 2} }}\\
= \dfrac{{\sqrt {10} – \sqrt 5 + \sqrt {14 + 3\sqrt {10} } }}{{2\sqrt 2 + 2 + \sqrt {11 – 2\sqrt 2 } }}
\end{array}\)