Сократите дробь:a) √5+1/√10+√2 б) 2-√2/√6-√3в) x²-2/√2x+2

a) [latex]\frac{\sqrt{5}+1}{\sqrt{10}+\sqrt{2}}=\frac{\sqrt{2}(\sqrt{5}+1)}{\sqrt{2}(\sqrt{10}+\sqrt{2})}=\frac{\sqrt{10}+\sqrt{2}}{\sqrt{2}(\sqrt{10}+\sqrt{2})}=\frac{1}{\sqrt{2}}[/latex] б) [latex]\frac{2-\sqrt{2}}{\sqrt{6}-\sqrt{3}}=\frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{3}(\sqrt{2}-1)}=\frac{\sqrt{2}}{\sqrt{3}}= \sqrt{ \frac{2}{3} }=ili =\frac{\sqrt{3}*\sqrt{2}}{\sqrt{3}*\sqrt{3}}=\frac{\sqrt{6}}{3}[/latex] в) [latex]\frac{x^2-2}{\sqrt{2}x+2}=\frac{(x-\sqrt{2})(x+\sqrt{2})}{\sqrt{2}(x+\sqrt{2})}=\frac{x-\sqrt{2}}{\sqrt{2}}[/latex]