解下列方程:(1)x2+3x2x2+2x−8+x2+x−43x2+9x=11/12;(2)3+(x-1998)3=1;(3)13x−x2x+1(x+13−xx+1)=42.
问题描述:
解下列方程:
(1)
+
x2+3x
2x2+2x−8
=
x2+x−4 3x2+9x
;11 12
(2)3+(x-1998)3=1;
(3)
(x+13x−x2
x+1
)=42. 13−x
x+1
答
(1)设
=y
x2+3x
x2+x-4
则原方程化为
y+1 2
=1 3y
11 12
解得x1=-1,x2=-4,x3,4=
5±
89
2
(2)设1999-x=a,x-1998=b,1999-x+x-1998=1,
则原方程a3+b3=(a+b)3得ab=0,即(x-1998)=0
∴x1=1999,x2=1998
(3)设y=
,则xy(x+y)=42,又xy+(x+y)=13-x x+1
+13x-x2
x+1
=13
x2+13 x+1
∴xy,x+y是方程x2-13x+42=0的两个根,
解得x1=6,x2=7,即
或
x+y=7 xy=6
x+y=6 xy=7
进而可得x1=1,x2=3+
,x3=3-
2
2