Đáp án:
$1)\\ a)max_A= 4 \Leftrightarrow x=1\\ b)max_B=9 \Leftrightarrow x=3\\ 2)\\ a)x(x+2)+y(y-2)-2xy+37 =100\\ b)x^2(x+1)-y^2(y-1)+xy-3xy(x-y+1)-95=297$
Giải thích các bước giải:
$1)\\ a)A=-x^2+2x+3\\ =-x^2+2x-1+4\\ =-(x-1)^4+4 \ge 4 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-1=0\Leftrightarrow x=1$
$b)B=6x-x^2\\ =-x^2+6x-9+9\\ =-(x-3)^2+9 \ge 9 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-3=0\Leftrightarrow x=3$
$2)\\ a)x(x+2)+y(y-2)-2xy+37\\ =x^2+2x+y^2-2y-2xy-2y\\ =x^2-2xy+y^2+2x-2y+37\\ =(x-y)^2+2(x-y)+37\\ =7^2+2.7+37\\ =100\\ b)x^2(x+1)-y^2(y-1)+xy-3xy(x-y+1)-95\\ =x^3+x^2-y^3+y^2+xy-3x^2y+3xy^2-3xy-95\\ =x^3-3x^2y+3xy^2-y^3+x^2+y^2-2xy-95\\ =(x-y)^3+(x-y)^2-95\\ =7^3+7^2-95\\ =297$
