\(x^2-2x\left(y+1\right)+\left(y+1\right)^2=-y^2+5y+1\)

\(\left(x-y-1\right)^2=-y^2+5y+1\ge0\)

y\(\in\){0;1;2;3;4;5}

+y=0 => (x-1)²=1=> x=0 hoặc x =2

+y =1 => (x-2)² =5 loại

+y=2 => (x-3)² =7 loại

+ y =3 => (x-4)² =7 loại

+y=4 => (x-5)² =5 loại

+ y=5 => (x-6)² = 1 => x =7 hoặc x=5

Vậy (x;y) là (0;0); (2;0) ;(5;5);(7;5).

Ta có :1990^10 + 1990^9 = 1990^9 . (1990+1) = 1990^9 . 1991 
và 1991^10 = 1991^9 . 1991 
Vì 1990 < 1991 nên 1990^10 + 1990^9 < 1991^10

\(\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\2+\dfrac{2}{x-1}-1-\dfrac{3}{y-1}=\dfrac{26}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x-2}+\dfrac{4}{y+1}=\dfrac{34}{5}\\\dfrac{6}{x-1}-\dfrac{9}{y-1}=\dfrac{63}{5}\end{matrix}\right.\)

\(\dfrac{4}{y+1}+\dfrac{9}{y-1}=-\dfrac{29}{5}=>y=....\)

Nếu đề là y+1 thì

\(\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{17}{5}\\2+\dfrac{2}{x-2}-1-\dfrac{1}{y+1}=\dfrac{26}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+2b=\dfrac{17}{5}\\2a-b=\dfrac{21}{5}\end{matrix}\right.\)