(X+1)$^{2}$ =x+1
4-x=2(4-x)$^{2}$
x(y+1)-y=1
($x^{2}$+1)(x-2)+2x=4
(X+1)$^{2}$ =x+1 4-x=2(4-x)$^{2}$ x(y+1)-y=1 ($x^{2}$+1)(x-2)+2x=4
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(X+1)$^{2}$ =x+1
4-x=2(4-x)$^{2}$
x(y+1)-y=1
($x^{2}$+1)(x-2)+2x=4
Acacia
1) x^2 + 2x + 1 = x + 1
x^2 + 2x + 1 – x – 1 = 0
x^2 + x = 0
x(x+1) = 0
Vậy x = 0 hoặc x=-1
2) (4 – x) – 2 (x-4)^2 = 0
(4-x) +2.(4-x)^2 =0
(4-x) [1+2.(4-x)] = 0
(4-x) (1+8-2x) = 0
(4-x) (9-2x) = 0
Vậy x = 4
3) x(y+1) – y – 1 = 0
x(y+1) – (y+1) = 0
(y+1) (x-1) = 0
Vậy y = -1 và x = 1
4) (x^2+1)(x-2) + 2x – 4 = 0
(x^2+1)(x-2) + 2(x – 2) = 0
(x-2) (x^2+1+2) = 0
(x-2) (x^2 +3) = 0
Vậy x=2
Jezebel
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