Question

解方程:(x²+x+1)/(x²+1)+(2x²+x+2)/(x²+x+1)=19/6

Answer 1

:(x²+x+1)/(x²+1)+(2x²+x+2)/(x²+x+1)=19/6
:(x²+x+1)/(x²+1)+(x²+x+1+x²+1)/(x²+x+1)=19/6
:(x²+x+1)/(x²+1)+1+ (x²+1)/(x²+x+1)=19/6
设:(x²+x+1)/(x²+1)=y
∴原方程可化为
y +1+1/y=19/6
6y²-13y+6=0
(3y-2)(2y-3)=0
∴y=2/3 y=3/2
∴:(x²+x+1)/(x²+1)=2/3 :(x²+x+1)/(x²+1)=3/2
2x²+2=3x²+3x+3 3x²+3=2x²+2x+2
x²+3x+1=0 x²-2x+1=0
x=(-3±√5)/2 x=1

Answer 2

Answer 3

(x²+x+1)/(x²+1)+(2x²+x+2)/(x²+x+1)=19/6
(x²+x+1)²/(x²+1)(x²+x+1)+(2x²+x+2)(x²+1)/(x²+x+1)(x²+1)=19/6
(x∧4+2x³+3x²+2x+1)+(2x∧4+x³+4x²+x+2)/(x²+1)(x²+x+1)=19/6
(3x∧4+3x³+7x²+3x+3)/(x∧4+x³+2x²+x+1)=19/6
3+x²/(x∧4+x³+2x²+x+1)=19/6
x/(x²+1)(x²+x+1)=1/6
∴x=1