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七年级数学solution一般
题目
定义一种新运算"☆",规则为:mm☆n=mn+mn−nn=m^{n}+mn-n,例如:22☆3=23+2×3−3=113=2^{3}+2\times 3-3=11.据此解答下列问题:
(1)(1)求(−2)\left(-2\right)☆44的值；
(2)(2)求(−1)\left(-1\right)☆[(−5)[\left(-5\right)☆2]2]的值.
知识点：一元一次不等式的应用章节：未标注

答案与解析

答案

(1)∵m\left(1\right)\because m☆n=mn+mn−nn=m^{n}+mn-n，
∴(−2)\therefore \left(-2\right)☆44
=(−2)4+(−2)×4−4=\left(-2\right)^{4}+\left(-2\right)\times 4-4
=16+(−8)+(−4)=16+\left(-8\right)+\left(-4\right)
=4=4；
(2)∵m(2)\because m☆n=mn+mn−nn=m^{n}+mn-n，
∴(−1)\therefore \left(-1\right)☆[(−5)[\left(-5\right)☆2]2]
=(−1)=\left(-1\right)☆[(−5)2+(−5)×2−2][\left(-5\right)^{2}+\left(-5\right)\times 2-2]
=(−1)=\left(-1\right)☆(25−10−2)\left(25-10-2\right)
=(−1)=\left(-1\right)☆1313
=(−1)13+(−1)×13−13=\left(-1\right)^{13}+\left(-1\right)\times 13-13
=(−1)+(−13)+(−13)=\left(-1\right)+\left(-13\right)+\left(-13\right)
=−27=-27.

解析

(1)∵m\left(1\right)\because m☆n=mn+mn−nn=m^{n}+mn-n，
∴(−2)\therefore \left(-2\right)☆44
=(−2)4+(−2)×4−4=\left(-2\right)^{4}+\left(-2\right)\times 4-4
=16+(−8)+(−4)=16+\left(-8\right)+\left(-4\right)
=4=4；
(2)∵m(2)\because m☆n=mn+mn−nn=m^{n}+mn-n，
∴(−1)\therefore \left(-1\right)☆[(−5)[\left(-5\right)☆2]2]
=(−1)=\left(-1\right)☆[(−5)2+(−5)×2−2][\left(-5\right)^{2}+\left(-5\right)\times 2-2]
=(−1)=\left(-1\right)☆(25−10−2)\left(25-10-2\right)
=(−1)=\left(-1\right)☆1313
=(−1)13+(−1)×13−13=\left(-1\right)^{13}+\left(-1\right)\times 13-13
=(−1)+(−13)+(−13)=\left(-1\right)+\left(-13\right)+\left(-13\right)
=−27=-27.

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