1) \(P=\left(\sqrt{3}-1\right)\dfrac{3+\sqrt{3}}{2\sqrt{3}}=\left(\sqrt{3}-1\right)\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{2\sqrt{3}}=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{2}=\dfrac{3-1}{2}=1\)
2) \(B=\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}=\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-\sqrt{28}+\sqrt{54}=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}=5\sqrt{6}\)3) \(A=\dfrac{4}{\sqrt{3}-1}-\dfrac{2}{\sqrt{2}+\sqrt{3}}-\sqrt{8}=\dfrac{4\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\dfrac{2\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}-\sqrt{8}=\dfrac{4\sqrt{3}+4}{2}-\dfrac{2\sqrt{2}-2\sqrt{3}}{-1}-2\sqrt{2}=2\sqrt{3}+2+2\sqrt{2}-2\sqrt{3}-2\sqrt{2}=2\)4) \(\dfrac{50-\sqrt{25}}{\sqrt{36}}=\dfrac{50-5}{6}=\dfrac{45}{6}=\dfrac{15}{2}\)
5) \(M=\dfrac{6}{2-\sqrt{3}}+\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{75}=\dfrac{6\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2-\sqrt{3}-5\sqrt{3}=12+6\sqrt{3}+2-\sqrt{3}-5\sqrt{3}=14\)6) \(\dfrac{2}{\sqrt{2}+2}+\dfrac{1}{3}.\sqrt{18}=\dfrac{2\left(\sqrt{2}-2\right)}{\left(\sqrt{2}+2\right)\left(\sqrt{2}-2\right)}+\dfrac{3\sqrt{2}}{3}=-\sqrt{2}+2+\sqrt{2}=2\)
7) \(P=\left(\dfrac{1}{2-\sqrt{3}}-\dfrac{1}{2+\sqrt{3}}\right).\dfrac{\sqrt{3}-1}{3-\sqrt{3}}=\dfrac{2+\sqrt{3}-2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}.\dfrac{\sqrt{3}-1}{\sqrt{3}\left(\sqrt{3}-1\right)}=2\sqrt{3}.\dfrac{1}{\sqrt{3}}=2\)8) \(A=\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{\left(\sqrt{2+\sqrt{3}}\right)^2-\left(\sqrt{2-\sqrt{3}}\right)^2}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}=\dfrac{2+\sqrt{3}-2+\sqrt{3}}{\sqrt{4-3}}=2\sqrt{3}\)
