Tính: $\frac{1}{√3}$ + $\frac{1}{3√2}$ + $\frac{1}{√3}$$\sqrt[]{\frac{5}{12}-\frac{1}{√6}}$
Tính: $\frac{1}{√3}$ + $\frac{1}{3√2}$ + $\frac{1}{√3}$$\sqrt[]{\frac{5}{12}-\frac{1}{√6}}$
By Melanie
By Melanie
Tính: $\frac{1}{√3}$ + $\frac{1}{3√2}$ + $\frac{1}{√3}$$\sqrt[]{\frac{5}{12}-\frac{1}{√6}}$
Đáp án:
\(\dfrac{{\sqrt 3 }}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
\dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }}.\sqrt {\dfrac{5}{{12}} – \dfrac{1}{{\sqrt 6 }}} \\
= \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }}.\sqrt {\dfrac{{5 – 2\sqrt 6 }}{{12}}} \\
= \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }}.\sqrt {\dfrac{{3 – 2.\sqrt 3 .\sqrt 2 + 2}}{{12}}} \\
= \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }}.\sqrt {\dfrac{{{{\left( {\sqrt 3 – \sqrt 2 } \right)}^2}}}{{12}}} \\
= \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }}.\dfrac{{\sqrt 3 – \sqrt 2 }}{{2\sqrt 3 }}\\
= \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{{\sqrt 3 – \sqrt 2 }}{6}\\
= \dfrac{{2\sqrt 3 + \sqrt 2 + \sqrt 3 – \sqrt 2 }}{6}\\
= \dfrac{{3\sqrt 3 }}{6} = \dfrac{{\sqrt 3 }}{2}
\end{array}\)
Đáp án:
Giải thích các bước giải:
Chúc bn hok tốt !