解无理、分式方程x/(x+3)·6/(x+3)=x/(x-3)+6/(x+3)3x-1/x+6x/(3x^2-1)=5√
1个回答
x/(x+3)·6/(x+3)=x/(x-3)+6/(x+3)
去分母：6x=x(x+3)+6(x-3)
x^2+3x-18=0
(x+6)(x-3)=0
x=-6,3(增根,舍去）
3x-1/x+6x/(3x^2-1)=5
去分母：3x^2(3x^2-1)-(3x^2-1)+6x^2=5x(3x^2-1)
9x^4-3x^2-3x^2+1+6x^2=15x^3-5x
9x^4-15x^3+5x+1=0
9x^4-9x^3-6x^3+6x-x+1=0
(x-1)(9x^3-6x^2-6x-1)=0
(x-1)(9x^3+3x^2-9x^2-3x-3x-1)=0
(x-1)((3x+1)(3x^2-3x-1)=0
x=1,-1/3,(3+√21)/6,(3-√21)/6
√[2x+2√(2x-1)]+√[2x-2√(2x-1)]=a(a为常数)
令p=√[2x+2√(2x-1)],q=√[2x-2√(2x-1)]
则p+q=a>=0,平方 p^2+q^2+2pq=a^2
p^2+q^2=4x
pq=2|x-1|
4x+4|x-1|=a^2,即x+|x-1|=a^2/4
x>=1时,有：2x-1=a^2/4,得：x=1/2+a^2/8
x
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