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)
13xx2
x+1
(x+
13−x
x+1
)=42


(1)设

x2+3x
x2+x-4
=y
则原方程化为
1
2
y+
1
3y
=
11
12

解得x1=-1,x2=-4,x3,4=
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=
13-x
x+1
,则xy(x+y)=42,又xy+(x+y)=
13x-x2
x+1
+
x2+13
x+1
=13
∴xy,x+y是方程x2-13x+42=0的两个根,
解得x1=6,x2=7,即
x+y=7
xy=6
x+y=6
xy=7

进而可得x1=1,x2=3+
2
,x3=3-
2