这四道题用换元法怎么做 要过程 10
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9,
=[(x+y)^3-1]+2xy(1-x-y)
=(x+y-1)[(x+y)^2+(x+y)+1]-2xy(x+y-1)
=(x+y-1)[(x+y)^2+(x+y)+1-2xy]
=(x+y-1)(x^2+y^2+x+y+1)
10,
=(a+b)^2-(2+2ab)(a+b)+4ab+(1-2ab+a^2*b^2)
=(a+b)^2-(2+2ab)(a+b)+(1+2ab+a^2*b^2)
=(a+b)^2-2(1+ab)(a+b)+(1+ab)^2
=(a+b-1-ab)^2
=(ab-a-b+1)^2
=[a*(b-1)-(b-1)]^2
=(a-1)^2*(b-1)^2
11,
上题的a和b分别换成x和y
所以=(x-1)^2*(y-1)^2
12,
=[(x+1)(x+6)][(x+2)(x+3)]+x^2
=(x^2+7x+6)(x^2+5x+6)+x^2
=[(x^2+6)+7x][(x^2+6)+5x]+x^2
=(x^2+6)^2+(7x+5x)(x^2+6)+35x^2+x^2
=(x^2+6)^2+12x*(x^2+6)+36x^2
=(x^2+6+6x)^2
=(x^2+6x+6)^2
=[(x+y)^3-1]+2xy(1-x-y)
=(x+y-1)[(x+y)^2+(x+y)+1]-2xy(x+y-1)
=(x+y-1)[(x+y)^2+(x+y)+1-2xy]
=(x+y-1)(x^2+y^2+x+y+1)
10,
=(a+b)^2-(2+2ab)(a+b)+4ab+(1-2ab+a^2*b^2)
=(a+b)^2-(2+2ab)(a+b)+(1+2ab+a^2*b^2)
=(a+b)^2-2(1+ab)(a+b)+(1+ab)^2
=(a+b-1-ab)^2
=(ab-a-b+1)^2
=[a*(b-1)-(b-1)]^2
=(a-1)^2*(b-1)^2
11,
上题的a和b分别换成x和y
所以=(x-1)^2*(y-1)^2
12,
=[(x+1)(x+6)][(x+2)(x+3)]+x^2
=(x^2+7x+6)(x^2+5x+6)+x^2
=[(x^2+6)+7x][(x^2+6)+5x]+x^2
=(x^2+6)^2+(7x+5x)(x^2+6)+35x^2+x^2
=(x^2+6)^2+12x*(x^2+6)+36x^2
=(x^2+6+6x)^2
=(x^2+6x+6)^2
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