Giải hệ phương trình sau: C x + 1 y ​ : C x y + 1 ​ : C x y − 1 ​ = 6 : 5 : 2

Giải thích

Điều kiện: ⎩ ⎨ ⎧ ​ 0 ≤ y ≤ x + 1 0 ≤ y + 1 ≤ x 0 ≤ y − 1 ≤ x x , y ∈ N ​ ⇔ ⎣ ⎡ ​ x , y ∈ N y ≥ 1 x ≥ y + 1 ​ Ta có: C x + 1 y ​ : C x y + 1 ​ : C x y − 1 ​ = 6 : 5 : 2 ⇔ 6 C x + 1 y ​ ​ = 5 C x y + 1 ​ ​ = 2 C x y − 1 ​ ​ ⇔ { 6 C x + 1 y ​ ​ = 5 C x y + 1 ​ ​ 5 C x y + 1 ​ ​ = 2 C x y − 1 ​ ​ ​ ⇔ { 6 1 ​ ⋅ y ! ( x + 1 − y )! ( x + 1 )! ​ = 5 1 ​ ⋅ ( y + 1 )! ( x − y − 1 )! x ! ​ 5 1 ​ ⋅ ( y + 1 )! ( x − y − 1 )! x ! ​ = 2 1 ​ ⋅ ( y − 1 )! ( x − y + 1 )! x ! ​ ​ ⇔ { 5 ( x + 1 ) ( y + 1 ) = 6 ( x − y ) ( x − y + 1 ) 2 ⋅ ( x − y ) ( x − y + 1 ) = 5 ⋅ y ⋅ ( y + 1 ) ​ ⇔ { 5 ( x + 1 ) ( y + 1 ) = 3 ⋅ 5 ⋅ y ⋅ ( y + 1 ) 2 ⋅ ( x − y ) ( x − y + 1 ) = 5 ⋅ y ⋅ ( y + 1 ) ​ ⇔ { x + 1 = 3 y 2 ⋅ ( x − y ) ( x − y + 1 ) = 5 ⋅ y ⋅ ( y + 1 ) ​ ⇔ { x = 3 y − 1 2 ⋅ ( 3 y − 1 − y ) ( 3 y − 1 − y + 1 ) = 5 ⋅ y ⋅ ( y + 1 ) ​ ⇔ { x = 3 y − 1 3 y 2 = 9 y ​ ⇔ { x = 3 y − 1 y = 3 ​ ⇔ { x = 8 y = 3 ​ Vậy nghiệm của hệ là: x = 8 , y = 3