Toán Lớp 7: Hay tìm GTNN của biểu thức sau a) A=|x-3|-4 b)B=(x-1)^2 +|y-5|-6 c)C=10-(x-4)^2 d) E=|x-7|+|3-x| g)F=|x-8|+|x-4|-2

Question

Toán Lớp 7:  Hay tìm GTNN của biểu thức sau a) A=|x-3|-4 b)B=(x-1)^2 +|y-5|-6 c)C=10-(x-4)^2 d) E=|x-7|+|3-x| g)F=|x-8|+|x-4|-2

tuyen98 · Answer

Giải đáp+Lời giải và giải thích chi tiết:     a,   A=|x-3|-4   |x-3|&ge;0&forall;x   &hArr;|x-3|-4&ge;-4   Dấu '=' xảy ra khi   |x-3|=0   &hArr;x-3=0   &hArr;x=3   Vậy A_{min}=-4 khi x=3   b,   B=(x-1)^2+|y-5|-6   $ begin{cases}(x-1)^2&ge;0&forall;x  |y-5|&ge;0&forall;y end{cases}$ &hArr;(x-1)^2+|y-5|-6&ge;-6   Dấu '=' xảy ra khi   $ begin{cases}(x-1)^2=0  |y-5|=0 end{cases}$ $&hArr; begin{cases}x-1=0  y-5=0 end{cases}$   $&hArr; begin{cases}x=1  y=5 end{cases}$   Vậy B_{min}=-6 khi x=1;y=5   c,   C=10-(x-4)^2   (x-4)^2&ge;0&forall;x   &hArr;-(x-4)^2&le;0   &hArr;10-(x-4)^2&le;10   Dấu '=' xảy ra khi   (x-4)^2=0   &hArr;x-4=0   &hArr;x=4   Vậy C_{max}=10&hArr;x=4   d,   E=|x-7|+|3-x|&ge;|x-7+3-x|   &hArr;E=|x-7|+|3-x|&ge;4   Dấu '=' xảy ra khi   (x-7)(3-x)&ge;0   $&hArr; left[ begin{matrix} begin{cases}x-7&ge;0  3-x&ge;0 end{cases}   begin{cases}x-7&le;0  3-x&le;0 end{cases} end{matrix} right.$ $&hArr; left[ begin{matrix} begin{cases}x&ge;7  x&le;3 end{cases}(L)   begin{cases}x&le;7  x&ge;3 end{cases} end{matrix} right.$   Vậy E_{min}=4 khi 3&le;x&le;7   g,   F=|x-8|+|x-4|-2   F=|x-8|+|4-x|-2&ge;|x-8+4-x|-2   &hArr;F=|x-8|+|4-x|-2&ge;4-2   &hArr;F=|x-8|+|4-x|-2&ge;2   Dấu '=' xảy ra khi   (x-8)(4-x)&ge;0   $&hArr; left[ begin{matrix} begin{cases}x-8&ge;0  4-x&ge;0 end{cases}   begin{cases}x-8&le;0  4-x&le;0 end{cases} end{matrix} right.$   $&hArr; left[ begin{matrix} begin{cases}x&ge;8  x&le;4 end{cases}(L)   begin{cases}x&le;8  x&ge;4 end{cases} end{matrix} right.$   Vậy F_{min}=2 khi 4&le;x&le;8

baoquyen · Answer

$  $   a,   A = |x-3|-4   V&igrave; |x-3| &ge;0&forall;x   -> |x-3|-4 &ge; -4 &forall;x   ->A &ge; -4 &forall;x   Dấu '=' xảy ra khi :   |x-3|=0 &harr;x-3=0 &harr;x=3   Vậy min A=-4 &harr;x=3   $  $   b,   B = (x-1)^2 +|y-5| - 6   V&igrave; (x-1)^2 &ge; 0&forall;x, |y-5| &ge;0&forall;y   -> (x-1)^2 + |y-5| - 6 &ge;-6&forall;x,y   -> B &ge;-6&forall;x,y   Dấu '=' xảy ra khi :   (x-1)^2=0, |y-5|=0   &harr; x-1=0,y-5=0   &harr;x=1,y=5   Vậy min B=-6 &harr;x=1,y=5   $  $   c,   C=10 - (x-4)^2   V&igrave; (x-4)^2 &ge;0&forall;x -> - (x-4)^2 &le;0&forall;x   -> 10-(x-4)^2 &le;10&forall;x   ->C &le;10&forall;x   Dấu '=' xảy ra khi :   (x-4)^2=0 &harr; x-4=0 &harr;x=4   Vậy max C=10 &harr;x=4   $  $   d,   E=|x-7| + |3-x|   &Aacute;p dụng BĐT |a|+|b| &ge; |a+b| c&oacute; :   |x-7| + |3-x| &ge; |x-7+3-x|=|-4|=4&forall;x   ->E &ge;4&forall;x   Dấu '=' xảy ra khi :   (x-7)(3-x)&ge; 0   TH1 :   x-7 &ge; 0,3-x &ge; 0   ->x &ge; 7, x&le;3   -> 7&le;x&le;3 (V&ocirc; l&iacute;)   TH2 :   x-7 &le; 0,3-x &le; 0   ->x &le; 7, x &ge;3   ->3 &le;x&le;7 (Lu&ocirc;n đ&uacute;ng)   Vậy min E = 4 &harr;3&le;x&le;7   $  $   g,   F = |x-8|+|x-4|-2   ->F = |x-8|+|4-x|-2   &Aacute;p dụng BĐT |a|+|b| &ge; |a+b| c&oacute; :   |x-8|+|4-x| &ge; |x-8+4-x| = |-4|=4&forall;x   -> |x-8|+|4-x|-2 &ge; 4-2=2&forall;x   ->F &ge;2&forall;x   Dấu '=' xảy ra khi :   (x-8)(4-x) &ge; 0   TH1 :   x-8 &ge; 0, 4-x &ge; 0   ->x &ge; 8,x&le;4   -> 8&le;x&le;4 (V&ocirc; l&iacute;)   TH2 :   x-8 &le; 0,4-x &le; 0   ->x &le; 8, x &ge; 4   -> 4 &le;x&le;8 (Lu&ocirc;n đ&uacute;ng)   Vậy min F=2 &harr; 4&le;x&le;8