Ответ:
дай лучший ответ пж
Объяснение:
[tex]\displaystyle\bf\\\frac{3}{2\sqrt{3}-3\sqrt{2} } =\frac{3\cdot(2\sqrt{3} +3\sqrt{2}) }{(2\sqrt{3}-3\sqrt{2})\cdot(2\sqrt{3} +3\sqrt{2}) } =\frac{3\cdot(2\sqrt{3} +3\sqrt{2}) }{(2\sqrt{3})^{2} -(3\sqrt{2})^{2} } =\\\\\\=\frac{3\cdot(2\sqrt{3} +3\sqrt{2}) }{12 -18 } =\frac{3\cdot(2\sqrt{3} +3\sqrt{2}) }{ -6 } =-\frac{2\sqrt{3} +3\sqrt{2} }{2 }[/tex]
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Ответ:
дай лучший ответ пж
Объяснение:
[tex]\displaystyle\bf\\\frac{3}{2\sqrt{3}-3\sqrt{2} } =\frac{3\cdot(2\sqrt{3} +3\sqrt{2}) }{(2\sqrt{3}-3\sqrt{2})\cdot(2\sqrt{3} +3\sqrt{2}) } =\frac{3\cdot(2\sqrt{3} +3\sqrt{2}) }{(2\sqrt{3})^{2} -(3\sqrt{2})^{2} } =\\\\\\=\frac{3\cdot(2\sqrt{3} +3\sqrt{2}) }{12 -18 } =\frac{3\cdot(2\sqrt{3} +3\sqrt{2}) }{ -6 } =-\frac{2\sqrt{3} +3\sqrt{2} }{2 }[/tex]