三、解答题(本题共 8 小题,共 66 分)
15. (本小题 12 分)计算.
(1)$(\sqrt{3}+\sqrt{6})+(\sqrt{6}-\sqrt{3})$;
(2)$\sqrt{\frac{1}{16}}+\vert -\sqrt{2}\vert +\sqrt[3]{-8}-(\sqrt{\frac{1}{4}})^2$;
(3)$\sqrt{81}+\sqrt[3]{-27}-\sqrt{(-2)^2}+\vert \sqrt{3}-2\vert$.
15. (本小题 12 分)计算.
(1)$(\sqrt{3}+\sqrt{6})+(\sqrt{6}-\sqrt{3})$;
(2)$\sqrt{\frac{1}{16}}+\vert -\sqrt{2}\vert +\sqrt[3]{-8}-(\sqrt{\frac{1}{4}})^2$;
(3)$\sqrt{81}+\sqrt[3]{-27}-\sqrt{(-2)^2}+\vert \sqrt{3}-2\vert$.
答案
(1)$2\sqrt{6}$;(2)$\sqrt{2}-2$;(3)$6-\sqrt{3}$
解析
(1)$(\sqrt{3}+\sqrt{6})+(\sqrt{6}-\sqrt{3})=\sqrt{3}+\sqrt{6}+\sqrt{6}-\sqrt{3}=2\sqrt{6}$
(2)$\sqrt{\frac{1}{16}}+\vert -\sqrt{2}\vert +\sqrt[3]{-8}-(\sqrt{\frac{1}{4}})^2=\frac{1}{4}+\sqrt{2}+(-2)-(\frac{1}{2})^2=\frac{1}{4}+\sqrt{2}-2-\frac{1}{4}=\sqrt{2}-2$
(3)$\sqrt{81}+\sqrt[3]{-27}-\sqrt{(-2)^2}+\vert \sqrt{3}-2\vert=9+(-3)-2+(2-\sqrt{3})=9-3-2+2-\sqrt{3}=6-\sqrt{3}$
(2)$\sqrt{\frac{1}{16}}+\vert -\sqrt{2}\vert +\sqrt[3]{-8}-(\sqrt{\frac{1}{4}})^2=\frac{1}{4}+\sqrt{2}+(-2)-(\frac{1}{2})^2=\frac{1}{4}+\sqrt{2}-2-\frac{1}{4}=\sqrt{2}-2$
(3)$\sqrt{81}+\sqrt[3]{-27}-\sqrt{(-2)^2}+\vert \sqrt{3}-2\vert=9+(-3)-2+(2-\sqrt{3})=9-3-2+2-\sqrt{3}=6-\sqrt{3}$
16. (本小题 16 分)求 $x$ 的值.
(1)$x^2-81=0$;
(2)$3x^3+81=0$;
(3)$(x-1)^2=4$;
(4)$(2x-1)^3=-8$.
(1)$x^2-81=0$;
(2)$3x^3+81=0$;
(3)$(x-1)^2=4$;
(4)$(2x-1)^3=-8$.
答案
(1)$\boxed{\pm 9}$
(2)$\boxed{-3}$
(3)$\boxed{3 \mathrm{或} -1}$
(4)$\boxed{-\frac{1}{2}}$
解析
(1)方程$x^2 - 81 = 0$,移项得$x^2 = 81$,解得$x = \pm 9$;
(2)方程$3x^3 + 81 = 0$,移项并化简得$x^3 = -27$,解得$x = -3$;
(3)方程$(x - 1)^2 = 4$,开平方得$x - 1 = \pm 2$,即$x = 1 \pm 2$,解得$x = 3$或$x = -1$;
(4)方程$(2x - 1)^3 = -8$,开立方得$2x - 1 = -2$,解得$x = -\frac{1}{2}$;
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