Vă rog mult!!!!!!!!!!!!!!!
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Răspuns:
A=
[tex]( \frac{1}{2 \sqrt{3} } + \frac{3}{4 \sqrt{3} } - \frac{5}{6 \sqrt{3} } ) \times \frac{3 \sqrt{3} }{10} [/tex]
[tex]( \frac{6}{12 \sqrt{3} } + \frac{9}{12 \sqrt{3} } - \frac{10}{12 \sqrt{3} } ) \times \frac{3 \sqrt{3} }{10} [/tex]
[tex] \frac{5}{12 \sqrt{3} } \times \frac{3 \sqrt{3} }{10} [/tex]
[tex] \frac{1}{8} [/tex]