①3a^5b^5-1/3ab②x^3-7x^2+12x③(x-y)^2(x+y)+(x+y)^2(y-x)④(x^2-3)^2-x(x^2-3)-2x^2⑤先化简再求值:(a+2b)^2+2(a+2b)(2a+b)+(2a+b)^2其中a=11/5 ,b= - 6/5⑥因为7*8*9*10+1=7*(7+1)*(7+2)*(7+3)+1=〔7*(7+3)〕*〔(7+1)*(

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①3a^5b^5-1/3ab②x^3-7x^2+12x③(x-y)^2(x+y)+(x+y)^2(y-x)④(x^2-3)^2-x(x^2-3)-2x^2⑤先化简再求值:(a+2b)^2+2(a+2b)(2a+b)+(2a+b)^2其中a=11/5 ,b= - 6/5⑥因为7*8*9*10+1=7*(7+1)*(7+2)*(7+3)+1=〔7*(7+3)〕*〔(7+1)*(

①3a^5b^5-1/3ab②x^3-7x^2+12x③(x-y)^2(x+y)+(x+y)^2(y-x)④(x^2-3)^2-x(x^2-3)-2x^2⑤先化简再求值:(a+2b)^2+2(a+2b)(2a+b)+(2a+b)^2其中a=11/5 ,b= - 6/5⑥因为7*8*9*10+1=7*(7+1)*(7+2)*(7+3)+1=〔7*(7+3)〕*〔(7+1)*(
①3a^5b^5-1/3ab
②x^3-7x^2+12x
③(x-y)^2(x+y)+(x+y)^2(y-x)
④(x^2-3)^2-x(x^2-3)-2x^2
⑤先化简再求值:(a+2b)^2+2(a+2b)(2a+b)+(2a+b)^2其中a=11/5 ,b= - 6/5
⑥因为7*8*9*10+1=7*(7+1)*(7+2)*(7+3)+1
=〔7*(7+3)〕*〔(7+1)*(7+2)〕+1
=(7^2+3*7)(7^2+3*7+2)+1
=(7^2+3*7)^2+2(7^2+3*)+1
=(7^2+3*7+1)^2
=71^2
∴7*8*9*10+1是一个平方数
求证:a(a+2)(a+4)(a+6)+16是一个平方数

①3a^5b^5-1/3ab②x^3-7x^2+12x③(x-y)^2(x+y)+(x+y)^2(y-x)④(x^2-3)^2-x(x^2-3)-2x^2⑤先化简再求值:(a+2b)^2+2(a+2b)(2a+b)+(2a+b)^2其中a=11/5 ,b= - 6/5⑥因为7*8*9*10+1=7*(7+1)*(7+2)*(7+3)+1=〔7*(7+3)〕*〔(7+1)*(
答:
①原式
=ab/3*(9a^4b^4-1)
=ab/3(3a^2b^2+1)(3a^2b^2-1)
=ab/3(3a^2b^2+1)(√3ab+1)(√3ab-1) 注,如果没学到无理数的话此步不要.
②原式
=x(x^2-7x+12)
=x(x-4)(x-3)
③原式
=(x-y)(x+y)(x-y-(x+y))
=-2y(x-y)(x+y)
④原式
=(x^2-3-2x)(x^2-3+x)
=(x^2+x-3)(x-3)(x+1)
⑤原式
=((a+2b)+(2a+b))^2
=(3a+3b)^2
=9(a+b)^2
=9*(11/5-6/5)^2
=9
⑥证明:
原式
=[a(a+6)][(a+2)(a+4)]+16
=(a^2+6a)(a^2+6a+8)+16
=(a^2+6a)^2+8(a^2+6a)+16
=(a^2+6a+4)^2
所以a(a+2)(a+4)(a+6)+16是完全平方数.

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