5) ĐK:................
(1)x7+xy6y14y8=0x7(y2)7+y6.(xy2)
(xy2)(f(x))+y6.(xy2)=0
(xy2)(f(x)+y6)=0  

Theo khai triển nhị thức Newton cho hằng đẳng thức bậc 7 thi f(x)>0 f(x)+y6>0

Do đó : x=y2

Xét phương trình thứ 2:  Đặt {3y+6=ay1=b

Ta có: {a+b=7a3b2=7{b=7a=a3(7a)2=7()

Cái phương trình () ko nhẩm được nghiệm nên có thể dùng công thức Cacnado.
Từ đó tìm ra a,b thế vào tìm yx.......


Đặt {a=x>0b=y0
Hệ {1+a2b2+ab=a21+a3b3=a2+3a3b
(2)(ab1)(a2b2+ab+1)=a2+3a3b2
a2(ab1)=a2+3a3b2
a3b+a2=1
a2(ab+1)=1()
(1)a+b.a2(ab+1)=a3
a+b=a3b=a3a
Thay vào () đc: a=1b=0
..................

avt xấu vầy ==" có ai nhờ đổi đâu nhỉ ==" –  Trần Trân 12-08-16 05:54 PM
2)ĐK:x65
pt2(x2x2)+(5x+6(x+2))+(7x+11(x+3))=0
(x+1)(x2)(215x+6+x+217x+11+x+3)=0
x=1orx=2(do(...)>0)
KL:....
4. ĐK:....
(3x27x+33x25x1)(x22x23x+4)>0
2x+43x27x+3+3x25x13x6x2+2+x23x+4>0
(x2)(2...+.....+3...+...)>0
x<2 k/h vs đk x...
1)đk:...
pt(2)(x+y1)(1xy)=0
x+y=1orxy=1
TH1:x+y=1
(1)tt;x2+3x4=x1(VN)
TH2:y=x1
(1)tt:x2+5x4=x3{x3x=5(t/m)
(x;y)=(5;4)

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