例:由27÷5=5...2 即27/5=5+2/5可以想到:(x^2-3x+2)÷(x+9)=x+2...110 即(x^2-3x+2)/(x+9)=x-12+110/x+9例:由27÷5=5...2 即27/5=5+2/5可以想到:(x^2-3x+2)÷(x+9)=x+2 ...110 即(x^2-3x+2)/(x+9)=x-12+110/x+9求:(x^2+2x+3)/(x+1)-(x^2-x+8
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例:由27÷5=5...2 即27/5=5+2/5可以想到:(x^2-3x+2)÷(x+9)=x+2...110 即(x^2-3x+2)/(x+9)=x-12+110/x+9例:由27÷5=5...2 即27/5=5+2/5可以想到:(x^2-3x+2)÷(x+9)=x+2 ...110 即(x^2-3x+2)/(x+9)=x-12+110/x+9求:(x^2+2x+3)/(x+1)-(x^2-x+8
例:由27÷5=5...2 即27/5=5+2/5可以想到:(x^2-3x+2)÷(x+9)=x+2...110 即(x^2-3x+2)/(x+9)=x-12+110/x+9
例:由27÷5=5...2 即27/5=5+2/5可以想到:(x^2-3x+2)÷(x+9)=x+2 ...110
即(x^2-3x+2)/(x+9)=x-12+110/x+9
求:(x^2+2x+3)/(x+1)-(x^2-x+8)/(x-2)
例:由27÷5=5...2 即27/5=5+2/5可以想到:(x^2-3x+2)÷(x+9)=x+2...110 即(x^2-3x+2)/(x+9)=x-12+110/x+9例:由27÷5=5...2 即27/5=5+2/5可以想到:(x^2-3x+2)÷(x+9)=x+2 ...110 即(x^2-3x+2)/(x+9)=x-12+110/x+9求:(x^2+2x+3)/(x+1)-(x^2-x+8
(x^2+2x+3)/(x+1)-(x^2-x+8)/(x-2)
=(x^2+2x+1+2)/(x+1)-(x^2-x-2+10)/(x-2)
=[(x+1)^2+2]/(x+1)-[(x-2)(x+1)+10]/(x-2)
=x+1+2/(x+1)-(x+1)-10/(x-2)
=2/(x+1)-10/(x-2)
=(2x-4-10x-10)/(x+1)(x-2)
=(-8x-14)/(x+1)(x-2)