Toán Lớp 8: TÌM GTNN: a, F= 4x^2 +15x+2 b, G= 8(a+2)^3 -(2a+1)^3 c, H= (X-1)(x+5)(x^2+4x+5)

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Toán Lớp 8:  TÌM GTNN: a, F= 4x^2 +15x+2 b, G= 8(a+2)^3 -(2a+1)^3 c,  H= (X-1)(x+5)(x^2+4x+5)

hatuanlan · Answer

Giải đáp+Lời giải và giải thích chi tiết:    a, F = 4x&sup2; + 15x + 2          = (4x&sup2; + 15x + $ frac{225}{15}$ ) - $ frac{193}{16}$           = (2x + $ frac{15}{4}$ )&sup2; - $ frac{193}{16}$    V&igrave; (2x + $ frac{15}{4}$ )&sup2; &ge; 0 với &forall;x   n&ecirc;n (2x + $ frac{15}{4}$ )&sup2; - $ frac{193}{16}$  &ge; - $ frac{193}{16}$    Vậy min F = - $ frac{193}{16}$  khi x = $ frac{-15}{8}$    b, G = 8.(a+2)&sup3; - (2a+1)&sup3;          = (2a + 4)&sup3; - (2a + 1)&sup3;         = (2a+4-2a-1).{ (2a+4)&sup2; + (2a+4).(2a+1) + (2a+1)&sup2; }         = 3.(4a&sup2; + 16a + 16 + 4a&sup2; + 8a + 4 + 4a&sup2; + 4a +1 )         = 3.(12a&sup2; + 28a + 21)         = 36a&sup2; + 84a + 63         = (36a&sup2; + 84a + 49 ) + 14         = (6a + 7)&sup2; + 14   V&igrave; (6a + 7)&sup2; &ge; 0 với &forall;x   n&ecirc;n (6a + 7)&sup2; + 14 &ge; 14    Vậy min G = 14 khi a =$ frac{-7}{6}$    c, H = (x-1).(x+5).(x&sup2; + 4x + 5)          = (x&sup2; + 4x - 5).(x&sup2; + 4x + 5)          = (x&sup2; + 4x)&sup2; - 25   V&igrave; (x&sup2; + 4x)&sup2; &ge; 0 với &forall;x    n&ecirc;n (x&sup2; + 4x)&sup2; - 25 &ge; -25    Vậy min H = -25 khi x = 0 hoặc x = -4

ngoclandiep · Answer

$a)F=4x^2+15x+2$ $F=4x^2+2.2x. dfrac{15}{4}+( dfrac{15}{4})^2- dfrac{193}{16}$ $F=(2x+ dfrac{15}{4})^2- dfrac{193}{16} ge- dfrac{193}{16}$ Dấu $'='$ xảy ra <=>$2x+ dfrac{15}{4}=0$ <=>$2x=- dfrac{15}{4}$ <=>$x=- dfrac{15}{8}$ Vậy $MinF=- dfrac{193}{16}$ khi $x=- dfrac{15}{8}$ $b)G=8(a+2)^3-(2a+1)^3$ $G=8(a^3+6a^2+12a+8)-(8a^3+12a^2+6a+1)$ $G=8a^3+48a^2+96a+64-8a^3-12a^2-6a-1$ $G=36a^2+90a+63$ $G=36a^2+2.6a. dfrac{15}{2}+( dfrac{15}{2})^2+ dfrac{27}{4}$ $G=(6a+ dfrac{15}{2})^2+ dfrac{27}{4} ge dfrac{27}{4}$ Dấu $'='$ xảy ra <=>$6a+ dfrac{15}{2}=0$ <=>$6a=- dfrac{15}{2}$ <=>$a=- dfrac{5}{4}$ Vậy $MinG= dfrac{27}{4}$ khi $a=- dfrac{5}{4}$ $c)H=(x-1)(x+5)(x^2+4x+5)$ $H=(x^2+5x-x-5)(x^2+4x+5)$ $H=(x^2+4x-5)(x^2+4x+5)$ $H=(x^2+4x)^2-25 ge-25$ Dấu $'='$ xảy ra <=>$x^2+4x=0$ <=>$x(x+4)=0$ <=>$ left[ begin{matrix} x=0 x+4=0 end{matrix} right.$ <=>$ left[ begin{matrix} x=0 x=-4 end{matrix} right.$ Vậy $MinH=-25$ khi $ left[ begin{matrix} x=0 x=-4 end{matrix} right.$