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

Question

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

maingocquynhnhu2k1 · Answer

giải th&iacute;ch c&aacute;c bước giải:     A=8x^2-4x+1/4 x^2+15   =(8x^2+1/4 x^2)-4x+15   =33/4 x^2-4x+15    =33/4(x^2-16/33 x+20/11)   =33/4(x^2-2 . 8/33 x+64/1089+1916/1089)   =33/4(x-8/33)^2+479/33   Do 33/4(x-8/33)^2 >=0&forall;x   &harr;️33/4(x-8/33)^2+479/33>=479/33&forall;x   Dấu = xảy ra khi :    33/4(x-8/33)^2=0   &harr;️(x-8/33)^2=0   &harr;️x-8/33=0   &harr;️x=8/33   Vậy  text{Min}_ text{A}=479/33&harr;️x=8/33

minhtamnguyet88 · Answer

Giải đáp:    $A_{min}= dfrac{479}{33}$   Lời giải và giải thích chi tiết:   $A=8x^2-4x+ dfrac{1}{4}x^2+15$   $= dfrac{33}{4}x^2-4x+15$   $= dfrac{33}{4} bigg{(}x^2- dfrac{16}{33}x+ dfrac{20}{11} bigg{)}$   $= dfrac{33}{4} bigg{(}x^2-2.x. dfrac{8}{33}+ dfrac{64}{1089}+ dfrac{1916}{1089} bigg{)}$   $= dfrac{33}{4}. bigg{(}x- dfrac{8}{33} bigg{)}^2+ dfrac{479}{33}$   Ta có: $ bigg{(}x- dfrac{8}{33} bigg{)}^2 ge 0&rArr; dfrac{33}{4}. bigg{(}x- dfrac{8}{33} bigg{)}^2 ge 0$   $&rArr; dfrac{33}{4}. bigg{(}x- dfrac{8}{33} bigg{)}^2+ dfrac{479}{33} ge  dfrac{479}{33}$   $&rArr;A ge  dfrac{479}{33}&rArr;A_{min}= dfrac{479}{33}$   D&acirc;́u '=' xảy ra khi: $ bigg{(}x- dfrac{8}{33} bigg{)}^2= 0$   $&rArr;x= dfrac{8}{33}$   V&acirc;̣y $A_{min}= dfrac{479}{33}$ khi $x= dfrac{8}{33}$.