Toán Lớp 9: x khác 0 tính min `A=8x^2-4x+1/(4x^2)+15`

Question

Toán Lớp 9:  x khác 0 tính min A=8x^2-4x+1/(4x^2)+15

cobelolen · Answer

Giải đáp:   $A=8x^2-4x+ dfrac{1}{4x^2}+15$     $=4x^2-4x+1+ dfrac{1}{4x^2}-2. dfrac{1}{2x}.2x+4x^2+16$   $=(2x-1)^2+( dfrac{1}{2x}-2x)^2+16$   Ta thấy :$(2x-1)^2&ge;0&forall;x$   $( dfrac{1}{2x}-2x)^2+16&ge;16&forall;x$   $&rArr;(2x-1)^2+( dfrac{1}{2x}-2x)^2+16&ge;16&forall;x$   Dấu bằng xảy ra &hArr;$ left  { {{2x-1=0}    { dfrac{1}{2x}-2x=0}}  right.$    $&hArr; left  { {{x= dfrac{1}{2}}    { dfrac{1}{2x}=2x}}  right.$   $&rArr;x=  dfrac{1}{2}$   Vậy Min A=$16&hArr;x=  dfrac{1}{2}$

maianh67 · Answer

A = 8x^2 - 4x + 1/(4x^2) + 15      = 4x^2 + 4x^2 - 4x + (1/(2x))^2 + 15      = 4x^2 - 4x + 1 + 4x^2 - 2 . 2x . 1/(2x) + (1/(2x))^2 + 16     = (2x - 1)^2 + (2x - 1/(2x))^2 + 16   Do   (2x - 1)^2 + (2x - 1/(2x))^2 &ge; 0   &rArr; A &ge; 16   text(Dấu bằng xảy ra khi) 2x - 1 = 0  v&agrave;  2x - 1/(2x) = 0   &rArr; x = 1/2   Vậy  min  A = 16  khi  x = 1/2