分式的计算题,能做几题做几题!1.1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+4x+3)2.[a/

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分式的计算题,
能做几题做几题!
1.1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+4x+3)
2.[a/(a-b)-a^2/(a^2-2ab+b^2)]÷[a/(a-b)+a^2/(b^2-a^2)]
3.x+2y/(x^2-y^2-z^2-2yz)-x+2y/(y^2+z^2-x^2+2yz)
4.b/(a-b+c)+2a+c/(b-a-c)+b-c/(b-a-c))

1.〔1/(x+1)(x+2)〕+〔1/(x+2)(x+3)〕+〔1/(x+1)(x+3)〕=〔x+3/(x+1)(x+2)(x+3)〕+〔x+1/(x+1)(x+2)(x+3)〕+〔x+2/(x+1)(x+2)(x+3)〕=〔3x+6/(x+1)(x+2)(x+3)〕=3/(x+1)(x+3)
2.=[1/(a-b)-a/(a^2-2ab+b^2)]÷[1/(a-b)+a/(b^2-a^2)] = [1/(a-b)-a/(a-b)^2]÷[1/(a-b)-a/(a-b)(a+b)]=[(a-b-a)/(a-b)^2]÷[(a+b-a)/(a-b)(a+b)]=
[(a-b-a)/(a-b)^2]÷[(a+b-a)/(a-b)(a+b)]=[(-b)/(a-b)^2]×[(a-b)(a+b)/(+b)]=-(a+b)/(a-b)
3.原式是否为：(x+2y)/(x^2-y^2-z^2-2yz)-(x+2y)/(y^2+z^2-x^2+2yz) =(x+2y)/(x^2-y^2-z^2-2yz)+(x+2y)/(x^2-y^2-z^2-2yz) =2(x+2y)/(x^2-y^2-z^2-2yz)+(x+2y)/(x^2-y^2-z^2-2yz) =2(x+2y)/(x+y+z)(x-y-z)
4.原式是否为：b/(a-b+c)+(2a+c)/(b-a-c)+(b-c)/(b-a-c)=b/(a-b+c)+(-2a-c)/(a-b+c)+(c-b)/(a-b+c)=(b-2a-c+c-b)/(a-b+c)=-2a/(a-b+c)

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