题目

（1）x^2+3x+2分之x-1+2+x-x^2分之6-4-x^2分之10-x
(2)x-y分之x乘以x+y分之y^2-x^4-y^4分之x^4y除以x^2+y^2分之x^2
(3)当x=-2时,求x+1分之1-x^2-1分之x+3除以x^2-2x+1分之x^2+4x+3
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提问时间：2020-12-04

答案

x^2+3x+2分之x-1+2+x-x^2分之6-4-x^2分之10-x
=(x-1)/(x²+3x+2)+6/(2+x-x²)-(10-x)/(4-x²)
=(x-1)/(x+1)(x+2)-6/(x-2)(x+1)-(x-10)/(x-2)(x+2)
=[(x-1)(x-2)-6(x+2)-(x-10)(x+1)]/(x+1)(x+2)(x-2)
=(x²-3x+2-6x-12-x²+9x+10)/(x+1)(x+2)(x-2)
=0/(x+1)(x+2)(x-2)
=0
(2)x-y分之x乘以x+y分之y^2-x^4-y^4分之x^4y除以x^2+y^2分之x^2
=x/(x-y)*y²/(x+y)*(x⁴y)/(x⁴-y⁴)÷x²/(x²+y²)
=x^5y³/[(x²-y²)(x²-y²)(x²+y²)]÷x²/(x²+y²)
=x³y³/(x²-y²)²
(3)当x=-2时,求x+1分之1-x^2-1分之x+3除以x^2-2x+1分之x^2+4x+3
x+1分之1-x^2-1分之x+3除以x^2-2x+1分之x^2+4x+3
=1/(x+1)-(x+3)/(x²-1)÷(x²+4x+3)/(x²-2x+1)
=1/(x+1)-(x+3)/(x-1)(x+1)÷(x+1)(x+3)/(x-1)²
=1/(x+1)-1/(x+1)²
=(x+1-1)/(x+1)²
=x/(x+1)²
=-2/(-2+1)²
=-2/1
=-2