Tìm giá trị lớn nhất, giá trị nhỏ nhất (nếu có thể):
d, \(D = -x^2 + 30x - 10\)
e, \(E = -2x^2 + 9x + 30\)
f, \(F = -5x^2 - 20x - 4\)
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\(A=x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)
Vậy GTNN A là 6 khi x - 2 = 0 <=> x = 2
\(B=\left(1-x\right)\left(3x-4\right)=3x-4-3x^2+4x=-3x^2+7x-4\)
\(=-3\left(x^2-\frac{7}{3}x+\frac{4}{3}\right)=-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{1}{36}\right)=-3\left(x-\frac{7}{6}\right)^2+\frac{1}{12}\ge\frac{1}{12}\)
\(=3\left(x-\frac{7}{6}\right)^2-\frac{1}{12}\le-\frac{1}{12}\)Vậy GTLN B là -1/12 khi x = 7/6
\(C=3x^2-9x+5=3\left(x^2-3x+\frac{5}{3}\right)=3\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{7}{12}\right)\)
\(=3\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\ge-\frac{7}{4}\)Vậy GTNN C là -7/4 khi x = 3/2
\(D=-2x^2+5x+2=-2\left(x^2-\frac{5}{2}x-1\right)=-2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{41}{16}\right)\)
\(=-2\left(x-\frac{5}{4}\right)^2+\frac{21}{8}\le\frac{21}{8}\)Vậy GTLN D là 21/8 khi x = 5/4
a) Đặt A = \(3x^2+6x+4\)
\(A=3\left(x^2+2x+1\right)+1\)
\(A=3\left(x+1\right)^2+1\)
Mà \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow3\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow A\ge1\)
Dấu "=" xảy ra khi : \(x+1=0\Leftrightarrow x=-1\)
Vậy Min A =1 khi x = -1
\(C=16x^2-8x+2024\)
\(\Rightarrow C=16x^2-8x+1+2023\)
\(\Rightarrow C=\left(4x-1\right)^2+2023\ge2023\left(\left(4x-1\right)^2\ge0\right)\)
\(\Rightarrow Min\left(C\right)=2023\)
\(D=-25x^2+50x-2023\)
\(\Rightarrow D=-\left(25x^2-50x+25\right)-1998\)
\(\Rightarrow D=-\left(5x-5\right)^2-1998\le1998\left(-\left(5x-5\right)^2\le0\right)\)
\(\Rightarrow Max\left(D\right)=1998\)
\(B=-x^2+20x+100=-\left(x^2-20x+100\right)+200=-\left(x-10\right)^2+200\le200\left(-\left(x-10\right)^2\le0\right)\)
\(\Rightarrow Max\left(B\right)=200\)
\(E=\left(2x-1\right)^2-\left(3x+2\right)\left(x-5\right)\)
\(\Rightarrow E=4x^2-4x+1-\left(3x^2-13x-10\right)\)
\(\Rightarrow E=4x^2-4x+1-3x^2+13x+10\)
\(\Rightarrow E=x^2+9x+11=x^2+9x+\dfrac{81}{4}-\dfrac{81}{4}+11\)
\(\Rightarrow E=\left(x+\dfrac{9}{2}\right)^2-\dfrac{37}{4}\ge-\dfrac{37}{4}\left(\left(x+\dfrac{9}{2}\right)^2\ge0\right)\)
\(\Rightarrow Min\left(E\right)=-\dfrac{37}{4}\)
\(F=\left(3x-5\right)^2-\left(3x+2\right)\left(4x-1\right)\)
\(\Rightarrow F=9x^2-30x+25-\left(12x^2+3x-2\right)\)
\(\Rightarrow F=-3x^2-33x+27=-3\left(x^2-10x+9\right)\)
\(\Rightarrow F=-3\left(x^2-10x+25\right)+48=-3\left(x-5\right)^2+48\le48\left(-3\left(x-5\right)^2\le0\right)\)
\(\Rightarrow Max\left(F\right)=48\)
\(A=x^2-6x+10=x^2-2\cdot x\cdot3+3^2+1=\left(x-3\right)^2+1\ge1\)
Vậy GTNN của A bằng 1. Dấu "=" xảy ra \(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
\(B=4x-x^2-5=-\left(x^2-2\cdot x\cdot2+2^2+1\right)=-\left(x-2\right)^2+1\le1\)
Vây GTLN của B bằng 1. Dấu "=" xảy ra \(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
\(C=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\ge4\)
Vậy GTNN của C bằng 4. Dấu '=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
\(D=x^2+x+1=x^2+2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy GTNN của D bằng 3/4. Dấu '=" xảy ra \(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)
Toàn bộ đều tìm Max :)
D = -x2 + 30x - 10
D = -( x2 - 30x + 225 ) + 215
D = -( x - 15 )2 + 215
-( x - 15 )2 ≤ 0 ∀ x => -( x - 15 )2 + 215 ≤ 215
Đẳng thức xảy ra <=> x - 15 = 0 => x = 15
=> MaxD = 215 <=> x = 15
E = -2x2 + 9x + 30
E = -2( x2 - 9/2x + 81/16 ) + 321/8
E = -2( x - 9/4 )2 + 321/8
-2( x - 9/4 )2 ≤ 0 ∀ x => -2( x - 9/4 )2 + 321/8 ≤ 321/8
Đẳng thức xảy ra <=> x - 9/4 = 0 => x = 9/4
=> MaxE = 321/8 <=> x = 9/4
F = -5x2 - 20x - 4
F = -5( x2 + 4x + 4 ) + 16
F = -5( x + 2 )2 + 16
-5( x + 2 )2 ≤ 0 ∀ x => -5( x + 2 )2 + 16 ≤ 16
Đẳng thức xảy ra <=> x + 2 = 0 => x = -2
=> MaxF = 16 <=> x = -2
d) \(D=-x^2+30x-10\)
\(D=-\left(x^2-30x+10\right)\)
\(D=\left(x^2-30x+225-215\right)\)
\(D=-\left(x-15\right)^2+215\le215\)
Max D = 215 \(\Leftrightarrow x=15\)
e) \(E=-2x^2+9x+30\)
\(E=-2\left(x^2-\frac{9}{2}x-15\right)\)
\(E=-2\left(x-\frac{9}{4}\right)^2+\frac{321}{8}\le\frac{321}{8}\)
Max \(E=\frac{321}{8}\Leftrightarrow x=\frac{9}{4}\)
f) \(F=-5x^2-20x-4\)
\(F=-5\left(x^2+4x+\frac{4}{5}\right)\)
\(F=-5\left(x^2+4x+4+\frac{16}{5}\right)\)
\(F=-5\left(x+2\right)^2-16\le-16\)
Max F = -16 \(\Leftrightarrow x=-2\)