Question

分式计算：x^2+2x+1分之x+1÷2x+2分之x+3

2.(2x+2y分之x-y)-(x^2-y^2分之x^2+y^2)
3.x^2+4x+4分之x^2-1÷(x+1)*x-1分之x^2+3x+2
4.(m+n)(m-n)-2(m-n)^2

Answer 1

x^2+2x+1分之x+1÷2x+2分之x+3
=(x+1)/(x+1)²×2(x+1)/(x+3)
=2/(x+3)

2.(2x+2y分之x-y)-(x^2-y^2分之x^2+y^2)
=(x-y)/2(x+y)-(x²+y²)/(x+y)(x-y)
=[(x-y)²-2(x²+y²)]/2(x+y)(x-y)]
=-(x²+2xy+y²)/2(x+y)(x-y)
=-(x+y)²/2(x+y)(x-y)
=(x+y)/(2y-2x)

3.x^2+4x+4分之x^2-1÷(x+1)*x-1分之x^2+3x+2
=(x+1)(x-1)/(x+2)²×1/(x+1)×(x+1)(x+2)/(x-1)
=(x+1)/(x+2)

4.(m+n)(m-n)-2(m-n)^2
=m²-n²-2(m²-2mn+n²）
=-m²+4mn-3n²

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