Minh xin tra loi bai cua ban @ unknown va @ Hung?

Bai cua @unknown

Giải hệ phương trình:

x(x+1) + (y+1)/y = 4

{

x³.y³ + x².y + x.y² + 1 = 4y³

Theo minh de phai nhu the nay

x(x+1) + (y+1)/y² = 4

{

x³.y³ + x².y + x.y² + 1 = 4y³

x^2 +x + 1 /y + 1/y^2 = 4

{

x^3 + x^2/y^2 + x / y + 1/y^3 = 4

(x^2 + 1 /y )+ ( x + 1/y^2 )= 4

{

x^2 (x + 1 / y^2) + 1 / y ( x + y^2 ) = 4

(x^2 + 1 /y )+ ( x + 1/y^2 )= 4

{

(x + 1 / y^2) ( x^2 + 1/y ) = 4

Toi day quen thuoc nha ban

Bai cua @ Hung

gjaj phuong trjnh

4*(can bac hai (x-1)-3)*x^2 +13(can bac hai (x+1)-8)*x - 4 *can bac hai (x-1) - 3 = 0

DK x > = 1

Ban bien doi dua ve duoc

x. [cbh ( x+1)] . [ 2. cbh ( x + 1 ) - 3]^2 + [ 2 cbh ( x-1 ) - 1]^2 = 0

<=> 2. cbh ( x + 1 ) = 3 va 2 cbh ( x-1 ) = 1

<=> x = 5 / 4 ( nhan)