a/ $u+v=15→v=5-u$
Thay $v=5-u$ vào pt $uv=4\\→u(5-u)=4\\↔-u^2+5u-4=0\\↔u^2-5u+4=0\\↔u^2-4u-u+4=0\\↔u(u-4)-(u-4)=0\\↔(u-1)(u-4)=0\\↔u-1=0\quad or\quad u-4=0\\↔u=1\quad or\quad u=4\\→v=4\quad or\quad v=1$
Vậy (u,v)={(1;4); (4;1)}
b/ $u^2+v^2=34\\↔(u-v)^2+2uv=15\\↔(u-v)^2+30=34\\↔(u-v)^2=4\\↔u-v=2\quad or\quad u-v=-2\\↔u=2+v\quad or\quad u=v-2$
Thay $u=v+2$ vào pt $uv=15\\→(v+2)v=15\\↔v^2+2v-15=0\\↔v^2+5v-3v-15=0\\↔v(v+5)-3(v+5)=0\\↔(v-3)(v+5)=0\\v-3=0\quad or\quad v+5=0\\↔v=3\quad or\quad v=-5\\→u=5\quad or\quad u=-3\\→(u,v)=\{(3;5); (-5;-3)\}$
Thay $u=v-2$ vào pt $uv=15\\→(v-2)v=15\\↔v^2-2v-15=0\\↔v^2-5v+3v-15=0\\↔v(v-5)+3(v-5)=0\\↔(v+3)(v-5)=0\\↔v+3=0\quad or\quad v-5=0\\↔v=-3\quad or\quad v=5\\→u=-5\quad or\quad v=3\\→(u,v)=\{(-3;-5);(5;3)\}$
Suy ra $(u,v)=\{(3;5);(-5;-3);(-3;-5);(5;3)\}$
