[tex]3)\\\\\frac{\sqrt{ 7 }-10 }{2\sqrt{3}} =\frac{\sqrt{ 7 }-10 }{2\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3} }=\frac{(\sqrt{7}-10)*\sqrt{3}}{2* \sqrt{3} *\sqrt{3}}=\frac{(\sqrt{7}-10)*\sqrt{3}}{2*3}=\frac{(\sqrt{7}-10)*\sqrt{3}}{6}\\\\4)\\\\\frac{2\sqrt{3}+3\sqrt{2}}{\sqrt{7}}=\frac{2\sqrt{3}+3\sqrt{2}}{\sqrt{7}}*\frac{\sqrt{7}}{\sqrt{7}}=\frac{(2\sqrt{3}+3\sqrt{2})*\sqrt{7}}{ \sqrt{7}*\sqrt{7}}=\frac{(2\sqrt{3}+3\sqrt{2})*\sqrt{7}}{ 7}\\\\5)\\\\\frac{\sqrt{3}}{1-\sqrt{2}}=\frac{\sqrt{3}}{1-\sqrt{2}}*\frac{1+\sqrt{2}}{1+\sqrt{2}}= \frac{\sqrt{3}(1+\sqrt{2})}{(\sqrt{1})^2-(\sqrt{2})^2}= \frac{\sqrt{3}(1+\sqrt{2})}{1-2}= -\sqrt{3}(1+\sqrt{2})[/tex]
[tex]6)\\\\\frac{2}{\sqrt{3}-\sqrt{5}}=\frac{2}{\sqrt{3}-\sqrt{5}}*\frac{ \sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}}=\frac{2(\sqrt{3}+\sqrt{5})}{(\sqrt{3})^2-(\sqrt{5})^2}\frac{2(\sqrt{3}+\sqrt{5})}{3-5}=\frac{\not{2}^1(\sqrt{3}+\sqrt{5})}{-\not{2}^1}= -(\sqrt{3}+\sqrt{5})\\\\7)\\\\\frac{\sqrt{7}+1}{2-\sqrt{5}}=\frac{\sqrt{7}+1}{2-\sqrt{5}}*\frac{2+\sqrt{5}}{2+\sqrt{5}}=\frac{(\sqrt{7}+1)(2+\sqrt{5})}{(\sqrt{2})^2-(\sqrt{5})^2}=\frac{(\sqrt{7}+1)(2+\sqrt{5})}{2-5}=\\\\\\=\frac{(\sqrt{7}+1)(2+\sqrt{5})}{-3}=-\frac{(\sqrt{7}+1)(2+\sqrt{5})}{ 3}[/tex]
[tex]8)\\\\\frac{\sqrt{12}-\sqrt{48}}{2-\sqrt{96}}=\frac{\sqrt{12}-\sqrt{48}}{2-\sqrt{96}}*\frac{2+\sqrt{96}}{2+\sqrt{96}}=\frac{(\sqrt{12}-\sqrt{48})*(2+\sqrt{96})}{ 2^2-(\sqrt{96})^2}=\frac{(\sqrt{12}-\sqrt{48})*(2+\sqrt{96})}{4-96}=\\\\=\frac{(\sqrt{12}-\sqrt{48})*(2+\sqrt{96})}{-92}=-\frac{(\sqrt{12}-\sqrt{48})*(2+\sqrt{96})}{92}\\\\stosujemy wzory:\\\\(a-b)(a+b)=a^2-b^2\\\\\sqrt{a}*\sqrt{a}=a\\\\(\sqrt{a})^2=a[/tex]
