Question

va rog frumos doresc rezolvarea acestor 3 exercitii sau cate puteti rezolva​

Answer 1

Răspuns:

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Answer 2

$10.\\a)~\sqrt{3}\Big(\sqrt{3}+\sqrt{2}\Big)-\sqrt{2}\Big(\sqrt{3} -\sqrt{2}\Big)= \\\\=\sqrt{3}\cdot \sqrt{3} +\sqrt{3}\cdot \sqrt{2} -\sqrt{2} \cdot \sqrt{3} +\sqrt{2} \cdot \sqrt{2} =\sqrt{9} +\sqrt{6} -\sqrt{6} +\sqrt{4} =\\\\=3+2=5$

$b)~2\sqrt{2} \Big(\sqrt{18}-\sqrt{50}\Big)+4\sqrt{12}:\sqrt{3}=2\sqrt{2} \Big(3\sqrt{2} -5\sqrt{2}\Big)+4\sqrt{4} =\\\\=2\sqrt{2} \cdot 3\sqrt{2} -2\sqrt{2} \cdot 5\sqrt{2} +4 \cdot 2=6\sqrt{4} -10\sqrt{4} +8=6 \cdot 2-10 \cdot 2+8=\\\\=12-20+8= 0$

$c)~\sqrt{3}\Big(\sqrt{2} +3\sqrt{3} \Big)-\Big(\sqrt{3}+\sqrt{8}\Big)\cdot 2\sqrt{2} =\\\\=\sqrt{3} \cdot \sqrt{2} +\sqrt{3} \cdot 3\sqrt{3} -\sqrt{3} \cdot 2\sqrt{2} -\sqrt{8} \cdot 2\sqrt{2}=\sqrt{6} +3\sqrt{9} -2\sqrt{6} -2\sqrt{16}=\\\\=-\sqrt{6} +3 \cdot 3-2 \cdot 4=-\sqrt{6} +9-8=-\sqrt{6} +1$

$d)~\Big(2\sqrt{6}+\sqrt{54}\Big):\sqrt{6}-\Big(8\sqrt{5} -\sqrt{45}\Big):\sqrt{5} =2+\sqrt{9} -8+\sqrt{9} =\\\\=2+3-8+3=0$

$11.\\a)~\sqrt{24}+\Big(2\sqrt{3}-3\sqrt{12}+\sqrt{27}\Big) \cdot \sqrt{2}=\\\\=2\sqrt{6} + 2\sqrt{3} \cdot \sqrt{2} -3\cdot 2\sqrt{3} \cdot \sqrt{2} +3\sqrt{3} \cdot \sqrt{2} =\\\\=2\sqrt{6} +2\sqrt{6} -6\sqrt{6} +3\sqrt{6} =\sqrt{6}$

$b)~\Big(\sqrt{10} \cdot \sqrt{90}:\sqrt{50} \Big)^2-\Big(\sqrt{90}-\sqrt{40} \Big)^2=\\\\=\Big(\sqrt{900}:\sqrt{50} \Big)^2-\Big(3\sqrt{10} -2\sqrt{10} \Big)^2 =\Big(\sqrt{18} \Big)^2-\Big(\sqrt{10}\Big)^2=18-10=8$

$c)~\Big[-\sqrt{75}-4\Big(\sqrt{3} -2\sqrt{3}\Big)\Big] \cdot (-1)^{20} =\Big(-5\sqrt{3} -4 \cdot \sqrt{3} +4 \cdot 2\sqrt{3} \Big) \cdot 1=\\\\=-5\sqrt{3} -4\sqrt{3} +8\sqrt{3} =-\sqrt{3}$

$\displaystyle d)~\Big(2 \cdot 10^{-1}+0,2 \cdot \sqrt{0,04} \Big):1,2=\Bigg(2 \cdot \frac{1}{10} +\frac{2}{10} \cdot \sqrt{\frac{4}{100} } \Bigg):\frac{12}{10} =\\\\=\Bigg(\frac{2}{10} +\frac{2}{10} \cdot \frac{\sqrt{4} }{\sqrt{100} } \Bigg)\cdot \frac{10}{12}=\Bigg(\frac{2}{10} +\frac{2}{10} \cdot \frac{2}{10} \Bigg) \cdot \frac{10}{12} =\Bigg(\frac{2}{10} +\frac{4}{100}\Bigg) \cdot \frac{10}{12} =\\\\=\frac{20+4}{100} \cdot \frac{10}{12}=\frac{24}{100} \cdot \frac{10}{12} = \frac{2}{10} =\frac{1}{5}$

$\displaystyle 12.\\a)~\sqrt{3} \cdot \Bigg(\sqrt{6} +\sqrt{\frac{2}{3} } \Bigg)=\sqrt{3} \cdot \Bigg(\sqrt{6}+ \frac{\sqrt{2} }{\sqrt{3} } \Bigg)=\sqrt{3} \cdot \sqrt{6} +\sqrt{3} \cdot \frac{\sqrt{2} }{\sqrt{3} } =\\\\=\sqrt{18} +\sqrt{2} =3\sqrt{2} +\sqrt{2} =4\sqrt{2}$

$\displaystyle b)~\Bigg(\sqrt{2} -\frac{1}{\sqrt{2} } \Bigg):\sqrt{2} =\Bigg(\sqrt{2} -\frac{1}{\sqrt{2} }\Bigg)\cdot \frac{1}{\sqrt{2} } =\sqrt{2} \cdot \frac{1}{\sqrt{2} }-\frac{1}{\sqrt{2} }\cdot \frac{1}{\sqrt{2} }=\\\\=\frac{\sqrt{2} }{\sqrt{2} }-\frac{1}{\sqrt{4} }=1-\frac{1}{2} =\frac{2-1}{2} =\frac{1}{2}$

$\displaystyle c)~\Bigg(\frac{1}{\sqrt{7} }-\sqrt{7} \Bigg) \cdot \sqrt{7} =\frac{1}{\sqrt{7} }\cdot \sqrt{7} -\sqrt{7} \cdot \sqrt{7} =\frac{\sqrt{7} }{\sqrt{7} } -\sqrt{49}=1-7=-6$

$\displaystyle d)~\Bigg(\sqrt{3}- \frac{2}{\sqrt{3} } \Bigg) \cdot \frac{5\sqrt{3} }{6} +\frac{1}{6}=\frac{3-2}{\sqrt{3} }\cdot \frac{5\sqrt{3} }{6} +\frac{1}{6} =\frac{1}{\sqrt{3} }\cdot \frac{5\sqrt{3} }{6} +\frac{1}{6} =\\\\=\frac{5\sqrt{3} }{6\sqrt{3} }+\frac{1}{6} =\frac{5\sqrt{3} +\sqrt{3} }{6\sqrt{3} } =\frac{6\sqrt{3} }{6\sqrt{3} }= 1$