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\(a.A=5x-x^2=-\left(x^2-5x\right)=-\left[x^2-2x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\left(\dfrac{5}{2}\right)^2\right]=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\)
Vì \(\left(x-\dfrac{5}{2}\right)^2\ge0\forall x\in R\Rightarrow-\left(x-\dfrac{5}{2}\right)^2\le0\forall x\in R\)
\(\Rightarrow-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)
\(\Rightarrow Max_A=\dfrac{25}{4}\Leftrightarrow x=\dfrac{5}{2}\)
\(b.B=x-x^2=-\left(x^2-x\right)=-\left(x^2-2x\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2\right)=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\right]=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\in R\Rightarrow-\left(x-\dfrac{1}{2}\right)^2\le0\forall x\in R\)
\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(\Rightarrow Max_B=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
\(c.C=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+2^2-7\right)=-\left(x-2\right)^2+7\)
Vì \(\left(x-2\right)^2\ge0\forall x\in R\Rightarrow-\left(x-2\right)^2\le0\forall x\in R\)
\(\Rightarrow-\left(x-2\right)^2+7\le7\)
\(\Rightarrow Max_B=7\Leftrightarrow x=2\)
\(d.D=-x^2+6x-11=-\left(x^2-6x+11\right)=-\left(x^2-6x+3^2+2\right)=-\left(x-3\right)^2-2\)
Vì \(\left(x-3\right)^2\ge0\forall x\in R\Rightarrow-\left(x-3\right)^2\le0\forall x\in R\)
\(\Rightarrow-\left(x-3\right)^2-2\le-2\)
\(\Rightarrow Max_D=-2\Leftrightarrow x=3\)
\(e.E=5-8x-x^2=-\left(x^2+8x-5\right)=-\left(x^2+8x+4^2-21\right)=-\left(x+4\right)^2+21\)
Vì \(\left(x+4\right)^2\ge0\forall x\in R\Rightarrow-\left(x+4\right)^2\le0\forall x\in R\)
\(\Rightarrow-\left(x+4\right)^2+21\le21\)
\(\Rightarrow Max_E=21\Leftrightarrow x=-4\)
\(A=5x-x^2=-\left(x^2-5x+\frac{25}{4}\right)+\frac{25}{4}=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\forall x\)
Dấu '' = '' xảy ra khi: \(x-\frac{5}{2}=0\Rightarrow x=\frac{5}{2}\)
Vậy \(MaxA=\frac{25}{4}\) khi \(x=\frac{5}{2}\)
\(B=x-x^2-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu '' = '' xảy ra khi: \(x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
Vậy \(MaxB=\frac{1}{4}\) khi \(x=\frac{1}{2}\)
\(C=4x-x^2+3=7-\left(4-4x+x^2\right)=7-\left(2-x\right)^2\le7\forall x\)
Dấu '' = '' xảy ra khi: \(2-x=0\Rightarrow x=2\)
Vậy \(MaxC=7\) khi \(x=2\)
GTNN :
B=4x2+4x+11
= (2x)2+2*x*2+22+7
=(2x+2)2+7>= 7
dấu ''='' sảy ra khi 2x+2=0
=> x = -1
vậy GTNN của biểu thức B là 7 tại x = -1
\(B=4x^2+4x+11\)
\(=4x^2+4x+1+10\)
\(=\left(2x+1\right)^2+10\ge10\)
Dau "=" xay ra <=> \(x=-\frac{1}{2}\)
Vay.....
a) A= x2 + 4x + 5
=x2+4x+4+1
=(x+2)2+1≥0+1=1
Dấu = khi x+2=0 <=>x=-2
Vậy Amin=1 khi x=-2
b) B= ( x+3 ) ( x-11 ) + 2016
=x2-8x-33+2016
=x2-8x+16+1967
=(x-4)2+1967≥0+1967=1967
Dấu = khi x-4=0 <=>x=4
Vậy Bmin=1967 <=>x=4
Bài 2:
a) D= 5 - 8x - x2
=-(x2+8x-5)
=21-x2+8x+16
=21-x2+4x+4x+16
=21-x(x+4)+4(x+4)
=21-(x+4)(x+4)
=21-(x+4)2≤0+21=21
Dấu = khi x+4=0 <=>x=-4
Bài 1:
c)C=x2+5x+8
=x2+5x+\(\left(\dfrac{5}{2}\right)^2\)+\(\dfrac{7}{4}\)
=\(\left(x+\dfrac{5}{2}\right)^2\)+\(\dfrac{7}{4}\)\(\ge\dfrac{7}{4}\)
Vậy \(C_{min}=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{5}{2}\)
Ta có : A = x2 - 4x + 1
=> A = x2 - 2.x.2 + 4 - 3
=> A = (x - 2)2 - 3
Mà : (x - 2)2 \(\ge0\forall x\in R\)
Nên : (x - 2)2 - 3 \(\ge-3\forall x\in R\)
Vậy GTNN của A là -3 khi x = 2
\(B=4x^2+4x+11=\left(2x\right)^2+2.2x.1+1+10=\left(2x+1\right)^2+10\)
Vì \(\left(2x+1\right)^2\ge0\Rightarrow B=\left(2x+1\right)^2+10\ge10\)
Dấu "=" xảy ra khi (2x+1)2=0 <=> 2x+1=0 <=> x=-1/2
Vậy gtnn của B là 10 khi x=-1/2
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\(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\)
Dấu "=" xảy ra khi x=0 hoặc x=-5
a) \(A=\left(x^2-2.2x+4\right)-3\)
\(A=\left(x-2\right)^2-3\ge-3\Leftrightarrow x=2\)
Vậy minA = -3 khi x = 2
b) \(B=4x^2+4x+11\)
\(B=\left(\left(2x\right)^2+2x.1+1\right)+10\)
\(B=\left(2x+1\right)^2+10\ge10\Leftrightarrow x=-\frac{1}{2}\)
Vậy min B = 10 khi x = -1/2
c) \(C=\left(x11\right)\left(x+3\right)\left(x+2\right)\left(x+6\right)\)
\(C=\left(x-1\right)\left(x+6\right)\left(x+3\right)\left(x+2\right)\)
\(C=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(C=\left(x^2+5x\right)^2-36\ge-36\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=0\end{matrix}\right.\)
Vậy MinC= -36 khi x =0 và x = -5
d) \(D=2x^2+y^2-2xy+2x-4y+9\)
\(D=y^2-2y\left(x+2\right)+\left(x+2\right)^2-x^2-4x-4+2x^2+2x+9\)
\(D=\left(y^2-y-x\right)^2+x^2-2x+5\)
\(D=\left(y^2-x-2\right)+\left(x-1\right)^2+4\ge4\Leftrightarrow\left[{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
Vậy min D = 4 khi x = 1 và y = 3
a) A = -x2 - 4x - 2 = -x2 - 4x - 4 + 2 = -( x2 + 4x + 4 ) + 2 = -( x + 2 )2 + 2
\(-\left(x+2\right)^2\le0\forall x\Rightarrow-\left(x+2\right)^2+2\le2\)
Dấu " = " xảy ra <=> x + 2 = 0 => x = -2
Vậy AMax = 2 , đạt được khi x = -2
b) -2x2 - 3x + 5 = -2( x2 + 1/5x + 9/16 ) + 49/8 = -2( x + 3/4 )2 + 49/8
\(-2\left(x+\frac{3}{4}\right)^2\le0\forall x\Rightarrow-2\left(x+\frac{3}{4}\right)^2+\frac{49}{8}\le\frac{49}{8}\)
Dấu " = " xảy ra <=> x + 3/4 = 0 => x = -3/4
Vậy BMax = 49/8 , đạt được khi x = -3/4
c) C = ( 2 - x )( x + 4 ) = -x2 - 2x + 8 = -x2 - 2x - 1 + 9 = -( x2 + 2x + 1 ) + 9 = -( x + 1 )2 + 9
\(-\left(x+1\right)^2\le0\forall x\Rightarrow-\left(x+1\right)^2+9\le9\)
Dấu " = " xảy ra <=> x + 1 = 0 => x = -1
Vậy CMax = 9, đạt được khi x = -1
d) D = 5 - 8x - x2 = -x2 - 8x - 16 + 21 = -( x2 + 8x + 16 ) + 21 = -( x + 4 )2 + 21
\(-\left(x+4\right)^2\le0\forall x\Rightarrow-\left(x+4\right)^2+21\le21\)
Dấu " = " xảy ra <=> x + 4 = 0 => x = -4
Vậy DMax = 21 , đạt được khi x = -4
e) E = -3x( x + 3 ) - 7 = -3x2 - 9x - 7 = -3( x2 + 3x + 9/4 ) - 1/4 = -3( x + 3/2 )2 - 1/4
\(-3\left(x+\frac{3}{2}\right)^2\le0\forall x\Rightarrow-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\le-\frac{1}{4}\)
Dấu " = " xảy ra <=> x + 3/2 = 0 => x = -3/2
Vậy EMax = -1/4 , đạt được khi x = -3/2
A=(x2-4x+4)-5=(x-2)2-5≥-5
Dau bang xay ra khi: x=2
Vay GTNN cua A=-5 khi x=2
B=(4x2+4x+1)+10=(2x+1)2+10≥10
Dau bang xay ra khi: x=-1/2
Vay GTNN cua B=10 khi x=-1/2
C=[(x-1)(x+6)].[(x+2)(x+3)]
= (x2+5x-6)(x2+5x+6)
Dat x2+5x=a => (a-6)(a+6)=a2-36≥-36
Dau bang xay ra khi : a=0 => x=0 hoac x=-5
Vay GTNN cua C=-36 khi x=0 hoac c=-5
D=-(x2+8x-5)
=> -D=x2+8x-5=(x2+8x+16)-21=(x+4)2-21
=> D= 21-(x+4)2≤21
Dau bang xay ra khi : x=-4
Vay GTLN cua D=21 khi x=-4
E=-(x2-4x-1)=-(x2-4x+4-5)=-(x-2)2+5=5-(x-2)2≤5
Dau bang xay ra khi : x=2
Vay GTLN cua E=5 khi x=2
\(A=x^2-4x+1\\ =x^2-4x+4-3\\ =\left(x^2-4x+4\right)-3\\ =\left(x-2\right)^2-3\\ \text{Do }\left(x-2\right)^2\ge0\forall x\\ \Rightarrow A=\left(x-2\right)^2-3\ge-3\forall x\\ \text{Dấu }"="\text{ xảy ra khi: }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{Vậy }A_{\left(Min\right)}=-3\text{ }khi\text{ }x=2\)
\(B=4x^2+4x+11\\ =4x^2+4x+1+10\\ =\left(4x^2+4x+1\right)+10\\ =\left(2x+1\right)^2+10\\ \text{Do }\left(2x+1\right)^2\ge0\forall x\\ \Rightarrow B=\left(2x+1\right)^2+10\ge10\forall x\\ \text{Dấu }"="\text{ xảy ra khi: }\\ \left(2x+1\right)^2=0\\ \Leftrightarrow2x+1=0\\ \Leftrightarrow2x=-1\\ \Leftrightarrow x=-\dfrac{1}{2}\\ \\ \text{Vậy }B_{\left(Min\right)}=10\text{ }khi\text{ }x=-\dfrac{1}{2}\)
\(C=\left(x-1\right)\left(x+3\right)\left(x+2\right)\left(x+6\right)\\ =\left(x^2-x+6x-6\right)\left(x^2+3x+2x+6\right)\\ =\left(x^2+5x-6\right)\left(x^2+5x+6\right)\\ =\left(x^2+5x\right)-36\\ \text{Do }\left(x^2+5x\right)^2\ge0\forall x\\ \Rightarrow C=\left(x^2+5x\right)^2-36\ge-36\forall x\\ \text{Dấu }"="\text{ xảy ra khi: }\\ \left(x^2+5x\right)^2=0\\ \Leftrightarrow x^2+5x=0\\ \Leftrightarrow x\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\\ \text{Vậy }C_{\left(Min\right)}=-36\text{ }khi\text{ }x=-0\text{ hoặc }x=-5\)
a)
\(A=5x-x^2\)
\(A=-x^2+5x\)
\(A=-\left(x^2-5x\right)\)
\(A=-\left(x^2-2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\left(\frac{5}{2}\right)^2\right)\)
\(A=-\left[\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\right]\)
\(A=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\)
\(A=\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\)
mà mũ chẵn luôn >= 0
\(\Rightarrow A\le\frac{25}{4}\)
Dấu '=" xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Vậy,.........
b)
\(B=x-x^2\)
\(B=-x^2+x\)
\(B=-\left(x^2-x\right)\)
\(B=-\left(x^2-2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right)\)
\(B=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)
\(B=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\)
\(B=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\)
mà ( x - 1/2 )2 luôn lớn hơn hoặc bằng 0 với mọi x
\(\Rightarrow B\le\frac{1}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
Vậy,..........