5(x^2+1)/(x+1)+6(x+1)/x^2+1=17
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5(x^2+1)/(x+1)+6(x+1)/x^2+1=17
5(x^2+1)/(x+1)+6(x+1)/x^2+1=17
5(x^2+1)/(x+1)+6(x+1)/x^2+1=17
设(x²+1)/(x+1)=y
∴5y+6/y=17
5y²-17y+6=0
(5y-2)(y-3)=0
∴y=2/5 y=3
(x²+1)/(x+1)=2/5 (x²+1)/(x+1)=3
5x²+5=2x+2 x²+1=3x+3
5x²-2x+3=0 x²-3x-2=0
无解 x=(3±√17)/2
答:第二项分母是x^2+1吧?
5(x^2+1)/(x+1)+6(x+1)/(x^2+1)=17
令m=(x^2+1)/(x+1),方程化为:
5m+6/m=17
5m^2-17m+6=0
(5m-2)(m-3)=0
m=2/5或者m=3
所以:
m=(x^2+1)/(x+1)=2/5
5x^2+5=2x+2
5x^2...
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答:第二项分母是x^2+1吧?
5(x^2+1)/(x+1)+6(x+1)/(x^2+1)=17
令m=(x^2+1)/(x+1),方程化为:
5m+6/m=17
5m^2-17m+6=0
(5m-2)(m-3)=0
m=2/5或者m=3
所以:
m=(x^2+1)/(x+1)=2/5
5x^2+5=2x+2
5x^2-2x+3=0,判别式=(-2)^2-4*5*3=-56<0,无实数解
m=(x^2+1)/(x+1)=3
x^2+1=3x+3
x^2-3x-2=0
x=[3±√(9+8)]/2
x=(3±√17)/2
经检验,都是原分式方程的根
收起
5(x^2+1)/(x+1)+6(x+1)/(x^2+1)=17
令 (x^2+1)/(x+1)=y
5y+6/y-17=0
5y^2-17y+6=0
(5y-2)(y-3)=0
y1=2/5
y2=3
(1) (x^2+1)/(x+1)=2/5
5x^2+5=2x+2
5x^2-2x+3=0 (无解)
(2) (x^2+1)/(x+1)=3
x^2+1=3x+3
x^2-3x+2=0
4x^2-12x+8=0
4x^2-12x+9=1
(2x-3)^2=1
。。。