Đáp án:
Ta có
`x^2y + xy^2 + x + y = 240`
`<=> xy(x + y) + (x + y) = 240`
`<=> (x + y)(xy + 1) = 240`
`<=> (x + y) . 12 = 240`
`<=> x + y = 20`
`-> x^2 + y^2 = (x + y)^2 - 2xy = 20^2 - 2.11 = 378`
`x^3 + y^3 = (x + y)[(x^2 + y^2) - xy] = 20.[378 - 11] = 7340`
`-> x^5 + y^5 = (x^2 + y^2)(x^3 + y^3) - y^2x^3 - x^2y^3`
`= 378 . 7340 - x^2y^2(x + y)`
`= 2774520 - 11^2 . 20`
`= 2774520 - 2420`
`= 277210`
b, `x^2 - 4xy + 5y^2 = 2(x - y)`
`<=> x^2 - 4xy + 5y^2 - 2x + 2y = 0`
`<=> (x - 4xy + 4y^2) + y^2 - 2x + 2y = 0`
`<=> (x - 2y)^2 - 2(x - 2y) + 1 + y^2 - 6y + 9 - 10 = 0`
`<=> (x - 2y - 1)^2 + (y - 3)^2 = 10`
Có `10 = (±1)^2 + (±3)^2`
Th1 : $\left \{ {{x - 2y - 1 = 1} \atop {y - 3 = 3}} \right.$ <=> $\left \{ {{x = 14} \atop {y = 6}} \right.$
Th2 : $\left \{ {{x - 2y - 1 = 1} \atop {y - 3 = -3}} \right.$ <=> $\left \{ {{x = 2} \atop {y = 0}} \right.$
Th3 : $\left \{ {{x - 2y - 1 = -1} \atop {y - 3 = 3}} \right.$ <=> $\left \{ {{x = 12} \atop {y = 6}} \right.$
Th4 : $\left \{ {{x - 2y - 1 = -1} \atop {y - 3 = -3}} \right.$ <=> $\left \{ {{x = 0} \atop {y = 0}} \right.$
Vậy các cặp `(x,y)` thõa mãn là `(0,0) ; (2,0) ; (12,6) ; (14,6)`
c, Ta có
`P = x^2(x^4 - 1)(x^2 + 2) + 1`
`= x^2(x^2 - 1)(x^2 + 1)(x^2 + 2) + 1`
`= [x^2(x^2 + 1)].[(x^2 - 1)(x^2 + 2)] + 1`
`= (x^4 + x^2)(x^4 - x^2 + 2x^2 - 2) + 1`
`= (x^4 + x^2)(x^4 + x^2 - 2) + 1`
`= (x^4 + x^2)^2 - 2(x^4 + x^2) + 1`
`= (x^4 + x^2 - 1)^2`
Do `x ∈ Z -> x^4 + x^2 - 1 ∈ Z -> đpcm`
Giải thích các bước giải:
