Giải các phương trình:
a/1+2axa=b2a2a2+x22ax
b/x2ab2b2=abac22bc2+xbc
c/ x2+1n22n12nx=xn
d/x+x21x2:1a2(1+ax)2(a+x)2=ab(ba)2
a/1+2axa=b2a2a2+x22ax
Điều kiện xa
1+2axa=b2a2a2+x22ax
1+2axa=b2a2(xa)2
(xa)22a(xa)=b2a2
a2+x22ax2ax+2a2=b2a2
x24ax+4a2b2=0
Δ=4a24a2+b2=b2{x1=2abx2=2a+b nhận vì thỏa mãn điều kiện.



b/x2ab2b2=abac22bc2+xbc()
ab2b2=b(a2b)
ac22bc2=c2(a2b)
mẫu chung: bc2(a2b)
Điều kiện cho các chữ a,b, c: b0,c0,a2b
()c2x2c(a2b)b(ab)=0
Δ=c2(a2b)2+4bc2(ab)=c2[a24ab+4b2+4ab4b2]=c2.a2
[x1=c(a2b)ac2c2=bcx2=c(a2b)+ac2c2=abc

c/ x2+1n22n12nx=xn
x2+1n(nx2)+1nx2=xn
Điều kiện: n0,x2n
Khử mẫu số: x2+1+n=x(nx2)(n1)x22x(n+1)=0
Δ=1+(n21)=n2 Nghiệm:
[x1=1x2=n+1n1

d/x+x21x2:1a2(1+ax)2(a+x)2=ab(ba)2
x(1+x)(1x)(1+x).(1+ax+a+x)(1+axax)1a2=ab(ba)2
Điều kiện:
{ab,a1x±1
x(1x).(1a2)(1x2)1a2=ab(ba)2
x(1x)=ab(ba)2x2+xab(ba)2=0
Δ=1+4ab(ba)2=(a+bab)2
[x1=1(a+bab)2=(aba)x2=1+(a+bab)2=(bab)

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