Bài 1Thực hiện phép tính
$\frac{6-6\sqrt[2]{6} }{\sqrt[2]{6} – 1}$ – $\frac{5}{1+\sqrt[2]{6} }$
Bài 2: Rút gọn các biểu thức sau
1)
$\frac{\sqrt[2]{15} – \sqrt[2]{12} }{\sqrt[2]{5} – 2 }$ + $\frac{2}{1-\sqrt[2]{3} }$
2)
$\frac{\sqrt[2]{18} – \sqrt[2]{6} }{\sqrt[2]{3} – 1 }$ – $\frac{6 -2 \sqrt[2]{6} }{2\sqrt[2]{6} -4 }$
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\dfrac{{6 – 6\sqrt 6 }}{{\sqrt 6 – 1}} – \dfrac{5}{{1 + \sqrt 6 }}\\
= \dfrac{{6.\left( {1 – \sqrt 6 } \right)}}{{\sqrt 6 – 1}} – \dfrac{{5.\left( {\sqrt 6 – 1} \right)}}{{\left( {\sqrt 6 – 1} \right)\left( {\sqrt 6 + 1} \right)}}\\
= – 6 – \dfrac{{5.\left( {\sqrt 6 – 1} \right)}}{{6 – 1}}\\
= – 6 – \left( {\sqrt 6 – 1} \right)\\
= – 5 – \sqrt 6 \\
2,\\
1,\\
\dfrac{{\sqrt {15} – \sqrt {12} }}{{\sqrt 5 – 2}} + \dfrac{2}{{1 – \sqrt 3 }}\\
= \dfrac{{\sqrt 3 .\left( {\sqrt 5 – \sqrt 4 } \right)}}{{\sqrt 5 – 2}} + \dfrac{{2.\left( {1 + \sqrt 3 } \right)}}{{\left( {1 – \sqrt 3 } \right)\left( {1 + \sqrt 3 } \right)}}\\
= \dfrac{{\sqrt 3 .\left( {\sqrt 5 – 2} \right)}}{{\sqrt 5 – 2}} + \dfrac{{2.\left( {1 + \sqrt 3 } \right)}}{{1 – 3}}\\
= \sqrt 3 – \left( {1 + \sqrt 3 } \right)\\
= – 1\\
2,\\
\dfrac{{\sqrt {18} – \sqrt 6 }}{{\sqrt 3 – 1}} – \dfrac{{6 – 2\sqrt 6 }}{{2\sqrt 6 – 4}}\\
= \dfrac{{\sqrt 6 \left( {\sqrt 3 – 1} \right)}}{{\sqrt 3 – 1}} – \dfrac{{\sqrt 6 \left( {\sqrt 6 – 2} \right)}}{{2.\left( {\sqrt 6 – 2} \right)}}\\
= \sqrt 6 – \dfrac{{\sqrt 6 }}{2}\\
= \dfrac{{\sqrt 6 }}{2}
\end{array}\)