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八年级数学解答题一般
题目
(1)(1)已知2a12a-1的平方根是±3\pm 3,a+3b1a+3b-1的算术平方根是44,求ab+7ab+7的立方根.
(2)(2)如果实数aa,bb,cc在数轴上的对应点的位置如图所示,化简:a2a+c+b33\sqrt{a^2}-|a+c|+\sqrt[3]{b^3}.
知识点:数轴、平方根、算术平方根、立方根章节:未标注

答案与解析

答案

(1)2a1\left(1\right)\because 2a-1的平方根是±3\pm 3
2a1=9\therefore 2a-1=9
a=5\therefore a=5
a+3b1\because a+3b-1的算术平方根是44
a+3b1=16\therefore a+3b-1=16
5+3b1=16\therefore 5+3b-1=16
b=4\therefore b=4
ab+7=5×4+7=27\therefore ab+7=5\times 4+7=27
27\because 27的立方根是33
ab+7\therefore ab+7的立方根是33
(2)(2)由数轴得,a<b<0a \lt b \lt 0c>0c \gt 0a>c|a| \gt |c|
a+c<0\therefore a+c \lt 0
a2a+c+b33\therefore \sqrt{{a}^{2}}-|a+c|+\sqrt[3]{{b}^{3}}
=a(ac)+b=-a-\left(-a-c\right)+b
=a+a+c+b=-a+a+c+b
=c+b=c+b.

解析

(1)2a1\left(1\right)\because 2a-1的平方根是±3\pm 3
2a1=9\therefore 2a-1=9
a=5\therefore a=5
a+3b1\because a+3b-1的算术平方根是44
a+3b1=16\therefore a+3b-1=16
5+3b1=16\therefore 5+3b-1=16
b=4\therefore b=4
ab+7=5×4+7=27\therefore ab+7=5\times 4+7=27
27\because 27的立方根是33
ab+7\therefore ab+7的立方根是33
(2)(2)由数轴得,a<b<0a \lt b \lt 0c>0c \gt 0a>c|a| \gt |c|
a+c<0\therefore a+c \lt 0
a2a+c+b33\therefore \sqrt{{a}^{2}}-|a+c|+\sqrt[3]{{b}^{3}}
=a(ac)+b=-a-\left(-a-c\right)+b
=a+a+c+b=-a+a+c+b
=c+b=c+b.

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