Toán Lớp 8: tìm Min: A=x^2-2xy+2y^2+2x-10y+17

Question

Toán Lớp 8:  tìm Min: A=x^2-2xy+2y^2+2x-10y+17

quynhthanhnhu · Answer

Giải đáp:       Lời giải và giải thích chi tiết:   $A$ = $x^{2}$ + $2y^{2}$ - $2xy$ + $2x$ - $10y$ + $17$   = $x^{2}$ + $y^{2}$ + $1$ - $2xy$ + $2x$ - $2y$ + $y^{2}$ - $8y$ + $16$   = ($x^{2}$ + $y^{2}$ + $1$ - $2xy$ + $2x$ - $2y$) + ($y^{2}$ - $8y$ + $16$)   = $(x-y+1)^{2}$ + $(y - 4)^{2}$ $ geq$ $0$ $với$ $mọi$ $x$   $Dấu$ $bằng$ $xảy$ $ra$    <=> $y$-$4$ =$0$ $v&agrave;$ $x$-$y$+$1$ $=0$ $= >$ $y$ $=$ $4$          $x$-$4$+$1$=$40$ $=>$ $x$=$3$

chauhoangmy98 · Answer

A=x&sup2; - 2xy + 2y&sup2; + 2x - 10y + 17       =x&sup2; -2xy + y&sup2; + y&sup2; -8y + 16 +2x -2y+1       =(x-y)&sup2; +(y-4)&sup2; +2(x-y) + 1       =(x-y)&sup2; + 2(x-y) + 1 + (y-4)&sup2;        =(x-y+1)&sup2; +(y-4)&sup2;       (x-y+1)&sup2; +(y-4)&sup2; &ge; 0       hay A &ge; 0       &rArr; $Min_A$=0        Dấu bằng xảy ra khi :       $ left  { {{y-4=0}  atop {x-y+1=0}}  right.$       $ left  { {{y=4}  atop {x=3}}  right.$