题霸题霸学习平台
登录开始使用
← 返回公开题库
八年级数学解答题一般
题目
计算.
(1)8+32−2(1)\sqrt{8}+\sqrt{32}-\sqrt{2}；
(2)(212−13)×6−48÷6(2)(2\sqrt{12}-\sqrt{\frac{1}{3}})×\sqrt{6}-\sqrt{48}÷\sqrt{6}；
(3)(23−1)(23+1)−(1−23)2(3)(2\sqrt{3}-1)(2\sqrt{3}+1)-{(1-2\sqrt{3})}^{2}；
(4)∣1−2∣−83+(π−3.14)0−(15)−1(4)|1-\sqrt{2}|-\sqrt[3]{8}+{(π-3.14)}^{0}-{(\frac{1}{5})}^{-1}.
知识点：二次根式的加减法、二次根式的混合运算章节：第2章 实数 / 2.3 二次根式

答案与解析

答案

（1）8+32−2\sqrt{8}+\sqrt{32}-\sqrt{2}
=22+42−2=2\sqrt{2}+4\sqrt{2}-\sqrt{2}
=52=5\sqrt{2}；
(2)(212−13)×6−48÷6(2)(2\sqrt{12}-\sqrt{\frac{1}{3}})×\sqrt{6}-\sqrt{48}÷\sqrt{6}
=272−2−8=2\sqrt{72}-\sqrt{2}-\sqrt{8}
=2×62−2−22=2×6\sqrt{2}-\sqrt{2}-2\sqrt{2}
=122−2−22=12\sqrt{2}-\sqrt{2}-2\sqrt{2}
=92=9\sqrt{2}；
(3)(23−1)(23+1)−(1−23)2(3)(2\sqrt{3}-1)(2\sqrt{3}+1)-{(1-2\sqrt{3})}^{2}
=(12−1)−(1−43+12)=\left(12-1\right)-(1-4\sqrt{3}+12)
=11−1+43−12=11-1+4\sqrt{3}-12
=43−2=4\sqrt{3}-2；
(4)∣1−2∣−83+(π−3.14)0−(15)−1(4)|1-\sqrt{2}|-\sqrt[3]{8}+{(π-3.14)}^{0}-{(\frac{1}{5})}^{-1}
=2−1−2+1−5=\sqrt{2}-1-2+1-5
=2−7=\sqrt{2}-7.

解析

（1）8+32−2\sqrt{8}+\sqrt{32}-\sqrt{2}
=22+42−2=2\sqrt{2}+4\sqrt{2}-\sqrt{2}
=52=5\sqrt{2}；
(2)(212−13)×6−48÷6(2)(2\sqrt{12}-\sqrt{\frac{1}{3}})×\sqrt{6}-\sqrt{48}÷\sqrt{6}
=272−2−8=2\sqrt{72}-\sqrt{2}-\sqrt{8}
=2×62−2−22=2×6\sqrt{2}-\sqrt{2}-2\sqrt{2}
=122−2−22=12\sqrt{2}-\sqrt{2}-2\sqrt{2}
=92=9\sqrt{2}；
(3)(23−1)(23+1)−(1−23)2(3)(2\sqrt{3}-1)(2\sqrt{3}+1)-{(1-2\sqrt{3})}^{2}
=(12−1)−(1−43+12)=\left(12-1\right)-(1-4\sqrt{3}+12)
=11−1+43−12=11-1+4\sqrt{3}-12
=43−2=4\sqrt{3}-2；
(4)∣1−2∣−83+(π−3.14)0−(15)−1(4)|1-\sqrt{2}|-\sqrt[3]{8}+{(π-3.14)}^{0}-{(\frac{1}{5})}^{-1}
=2−1−2+1−5=\sqrt{2}-1-2+1-5
=2−7=\sqrt{2}-7.

AI 自由组卷

围绕这道题再组一份练习 →

完整试卷

浏览同年级试卷结构 →