17. 计算:
(1)$|\sqrt{5}-2|+\sqrt{9}+\sqrt{(-2)^{2}}-\sqrt[3]{-27}$.
(2)$2\sqrt{2}+\sqrt{9}+\sqrt[3]{-8}+|\sqrt{2}-2|$.
(1)$|\sqrt{5}-2|+\sqrt{9}+\sqrt{(-2)^{2}}-\sqrt[3]{-27}$.
(2)$2\sqrt{2}+\sqrt{9}+\sqrt[3]{-8}+|\sqrt{2}-2|$.
答案:
解:
(1)原式=$\sqrt{5}+6$.
(2)原式=$\sqrt{2}+3$.
(1)原式=$\sqrt{5}+6$.
(2)原式=$\sqrt{2}+3$.
证明:$\because DE// BC$,$\therefore \angle ABC= \angle ADE$(
$\because BE$,$DF分别是\angle ABC$,$\angle ADE$的平分线,
$\therefore \angle 3= \frac{1}{2}\angle ABC$,$\angle 4= \frac{1}{2}\angle ADE$.
$\therefore \angle 3= \angle 4$(
$\therefore$
两直线平行,同位角相等
).$\because BE$,$DF分别是\angle ABC$,$\angle ADE$的平分线,
$\therefore \angle 3= \frac{1}{2}\angle ABC$,$\angle 4= \frac{1}{2}\angle ADE$.
$\therefore \angle 3= \angle 4$(
等量代换
).$\therefore$
BE
$//$DF
(同位角相等,两直线平行
). $\therefore \angle 1= \angle 2$(两直线平行,内错角相等
).
答案:
解:两直线平行,同位角相等 等量代换
BE DF 同位角相等,两直线平行
两直线平行,内错角相等
BE DF 同位角相等,两直线平行
两直线平行,内错角相等
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