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七年级数学解答题一般
题目
已知:a=3|a|=3,b=2|b|=2,根据下列条件求值.
(1)(1)a+b>0a+b \gt 0,求a+ba+b的值;
(2)(2)a<ba \lt b,求abab的值;
(3)(3)ab<0ab \lt 0,求(ab)3\left(a-b\right)^{3}的值.
知识点:绝对值(二)、有理数的加法、有理数的乘法、代数式求值章节:第1章 有理数 / 1.2 有理数 / 1.2.4 绝对值

答案与解析

答案

(1)a=3\left(1\right)\because |a|=3b=2|b|=2
a=±3\therefore a=\pm 3b=±2b=\pm 2
a+b>0\because a+b \gt 0
a=3\therefore a=3b=2b=2a=3a=3b=2b=-2
a+b=3+2=5\therefore a+b=3+2=5a+b=3+(2)=1a+b=3+\left(-2\right)=1
(2)a<b(2)\because a \lt b
a=3\therefore a=-3b=±2b=\pm 2
b=2b=2时,ab=3×2=6ab=-3\times 2=-6
b=2b=-2时,ab=3×(2)=6ab=-3\times \left(-2\right)=6
ab=±6\therefore ab=\pm 6
(3)ab<0(3)\because ab \lt 0
a=3\therefore a=-3b=2b=2a=3a=3b=2b=-2
a=3a=-3b=2b=2时,
(ab)3=(32)3=125(a-b)^{3}=\left(-3-2\right)^{3}=-125
a=3a=3b=2b=-2时,
(ab)3=[3(2)]3=125(a-b)^{3}=\left[3-\left(-2\right)\right]^{3}=125
(ab)3=±125\therefore \left(a-b\right)^{3}=\pm 125.

解析

(1)a=3\left(1\right)\because |a|=3b=2|b|=2
a=±3\therefore a=\pm 3b=±2b=\pm 2
a+b>0\because a+b \gt 0
a=3\therefore a=3b=2b=2a=3a=3b=2b=-2
a+b=3+2=5\therefore a+b=3+2=5a+b=3+(2)=1a+b=3+\left(-2\right)=1
(2)a<b(2)\because a \lt b
a=3\therefore a=-3b=±2b=\pm 2
b=2b=2时,ab=3×2=6ab=-3\times 2=-6
b=2b=-2时,ab=3×(2)=6ab=-3\times \left(-2\right)=6
ab=±6\therefore ab=\pm 6
(3)ab<0(3)\because ab \lt 0
a=3\therefore a=-3b=2b=2a=3a=3b=2b=-2
a=3a=-3b=2b=2时,
(ab)3=(32)3=125(a-b)^{3}=\left(-3-2\right)^{3}=-125
a=3a=3b=2b=-2时,
(ab)3=[3(2)]3=125(a-b)^{3}=\left[3-\left(-2\right)\right]^{3}=125
(ab)3=±125\therefore \left(a-b\right)^{3}=\pm 125.

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