三、解答题(本大题共8小题,共72分.解答应写出过程)
17. (8分)计算.
(1)$\sqrt{4} + \sqrt[3]{-8} - \sqrt{\frac{1}{4}}$;
(2)$3\sqrt{2} - |\sqrt{3} - \sqrt{2}|$.

17. 【点拨】本题考查平方根,立方根,绝对值的计算.
【解析】(1)$\sqrt{4} + \sqrt[3]{-8} - \sqrt{\dfrac{1}{4}}$
$=2 + (-2) - \dfrac{1}{2}$
$= -\dfrac{1}{2}$.
(2)$3\sqrt{2} - |\sqrt{3} - \sqrt{2}|$
$=3\sqrt{2} -\sqrt{3} +\sqrt{2}$
$=4\sqrt{2} -\sqrt{3}$.

18. (8分)求下列各式中x的值:
(1)$x^{3} - 3 = \frac{3}{8}$;
(2)$(x - 1)^{2} = 25$.

18. 【点拨】本题考查利用立方根和平方根解方程.
【解析】(1)$x^3 - 3 = \dfrac{3}{8}$
$x^3 = \dfrac{3}{8} + 3$
$x^3 = \dfrac{27}{8}$
$x = \dfrac{3}{2}$.
(2)$(x-1)^2 = 25$
$x-1 = \pm5$,
$x=6$或$x=-4$.

19. (8分)如图，$∠ 1 + ∠ 2 = 180°, BF // DE$.
(1)求证: $∠ AGF = ∠ ABC$；
(2)若$DE ⊥ AC, ∠ 2 = 144°$, 求$∠ AFG$的度数.

19. 【点拨】本题考查平行线的判定与性质.
【解析】(1)证明:$\because BF// DE$,$\therefore ∠ 2 + ∠ DBF = 180°$.
$\because ∠ 1 + ∠ 2 = 180°$,$\therefore ∠ 1 = ∠ DBF$,
$\therefore BC// GF$,$\therefore ∠ AGF = ∠ ABC$.
(2)$\because DE⊥ AC$,$BF// DE$,$\therefore BF⊥ AC$,$\therefore ∠ AFB = ∠ AED = 90°$.
$\because ∠ 1 + ∠ 2 = 180°$,$∠ 2 = 144°$,$\therefore ∠ 1 = 36°$.
$\because ∠ AFG + ∠ 1 = ∠ AFB = 90°$,$\therefore ∠ AFG = 54°$.

20. (8分)看图填空:如图，$CF ⊥ AB$于点$F$，$DE ⊥ AB$于点$E$，$∠ 1 + ∠ EDC = 180°$，求证:$FG // BC$.
证明:$\because CF ⊥ AB, DE ⊥ AB$(已知),
$\therefore ∠ BED = ∠ BFC = 90°$(垂直的定义),
$\therefore ED // FC$(),
$\therefore ∠ 2 = ∠ 3$().
$\because ∠ 1 + ∠ EDC = 180°$(已知),$∠ 2 + ∠ EDC = 180°$(平角的定义),
$\therefore ∠ 1 = ∠ 2$(),
$\therefore ∠ 1 = ∠ 3$(等量代换),
$\therefore FG // BC$().

20. 【点拨】本题考查垂直的定义,平角的定义,平行线的判定与性质,掌握平行线的判定与性质是解题的关键.
【解析】证明:$\because CF ⊥ AB,DE ⊥ AB$(已知),
$\therefore ∠ BED = ∠ BFC = 90°$(垂直的定义),
$\therefore ED// FC$(同位角相等,两直线平行),
$\therefore ∠ 2 = ∠ 3$(两直线平行,同位角相等).
$\because ∠ 1 + ∠ EDC = 180°$(已知),$∠ 2 + ∠ EDC = 180°$(平角的定义),
$\therefore ∠ 1 = ∠ 2$(同角的补角相等),
$\therefore ∠ 1 = ∠ 3$(等量代换),
$\therefore FG// BC$(内错角相等,两直线平行).