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Bài 1:
a) A= x2 + 4x + 5
=x2+4x+4+1
=(x+2)2+1\(\ge\)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\(\ge\)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\(\le\)0+21=21
Dấu = khi x+4=0 <=>x=-4
b)đề sai à
ài 1:
a) A= x2 + 4x + 5
=x2+4x+4+1
=(x+2)2+1$\ge$≥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$\ge$≥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$\le$≤0+21=21
Dấu = khi x+4=0 <=>x=-4
b)đề sai à
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
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 - 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
Bài 5:
a/A = x2 - 6x + 10 = x2 - 6x + 9 + 1 = ( x - 3 )2 +1
Vì ( x - 3 )2 \(\ge\)0 nên ( x - 3 )2 + 1 \(\ge\)1
Giá trị nhỏ nhất của A là 1
b/ B = x ( x + 6 ) = x2 + 6x + 9 - 9 = ( x + 3 )2 - 9
Vì ( x + 3 )\(\ge\)0 nên ( x + 3 ) - 9\(\ge\)- 9
Giá trị nhỏ nhất của B là - 9
5 - A\(=x^2-6x+10\)
A\(=x^2-3x-3x+9+1\)
A\(=x\left(x-3\right)-3\left(x-3\right)+1\)
A\(=\left(x-3\right)\left(x-3\right)+1\)
A\(=\left(x-3\right)^2+1\)
Vì \(^{\left(x-3\right)^2\ge0\forall x}\)
\(\rightarrow\left(x-3\right)^2+1\ge1\forall x\)
Hay A\(\ge1\forall x\)
Dấu '' = '' xảy ra\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
B\(=x\left(x+6\right)\)
B\(=x^2+6x\)
B\(=x\left(x+3\right)+3\left(x+3\right)-9\)
B\(=\left(x+3\right)\left(x+3\right)-9\)
B\(=\left(x+3\right)^2-9\)
Vì\(\left(x+3\right)^2\ge0\forall x\)
\(\rightarrow\left(x+3\right)^2-9\ge-9\forall x\)
Hay B\(\ge-9\forall x\)
Dấu ''='' xảy ra \(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
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 \(C_{min}=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{5}{2}\)
\(1,a,A=x^2-6x+25\)
\(=x^2-2.x.3+9-9+25\)
\(=\left(x-3\right)^2+16\)
Ta có :
\(\left(x-3\right)^2\ge0\)Với mọi x
\(\Rightarrow\left(x-3\right)^2+16\ge16\)
Hay \(A\ge16\)
\(\Rightarrow A_{min}=16\)
\(\Leftrightarrow x=3\)
\(A=5x-x^2=-\left(x^2-5x\right)=-\left[x^2-2.x.\frac{5}{2}+\left(\frac{5}{2}\right)^2-\left(\frac{5}{2}\right)^2\right]=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\)
Vì \(\left(x-\frac{5}{2}\right)^2\ge0\left(x\in R\right)\)
nên \(-\left(x-\frac{5}{2}\right)^2\le0\left(x\in R\right)\)
do đó \(-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\left(x\in R\right)\)
Vậy \(Max_A=\frac{25}{4}\)khi \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
\(B=x-x^2=-\left(x^2-x\right)=-\left(x^2-2x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right)=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]=-\left(x-\frac{1}{2}^2\right)+\frac{1}{4}\)
Vì \(\left(x-\frac{1}{2}\right)^2\ge0\left(x\in R\right)\)
nên \(-\left(x-\frac{1}{2}\right)^2\le0\left(x\in R\right)\)
do đó \(-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\left(x\in R\right)\)
Vậy \(Max_B=\frac{1}{4}\)khi \(x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
\(C=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-2.x.2+2^2-7\right)=-\left(x-2\right)^2+7\)
Vì \(\left(x-2\right)^2\ge0\left(x\in R\right)\)
nên \(-\left(x-2\right)^2\le0\left(x\in R\right)\)
do đó \(-\left(x-2\right)^2+7\le7\left(x\in R\right)\)
Vậy \(Max_C=7\)khi \(x-2=0\Leftrightarrow x=2\)
\(D=-x^2+6x-11=-\left(x^2-6x+11\right)=-\left(x^2-2.x.3+3^2+2\right)=-\left(x-3^2\right)-2\)
Vì \(\left(x-3\right)^2\ge0\left(x\in R\right)\)
nên \(-\left(x-3\right)^2\le0\left(x\in R\right)\)
do đó \(-\left(x-3\right)^2-2\le-2\left(x\in R\right)\)
Vậy \(Max_D=-2\)khi \(x-3=0\Leftrightarrow x=3\)
\(E=5-8x-x^2=-\left(x^2+8x-5\right)=-\left(x^2+2.x.4+4^2-21\right)=-\left(x+4\right)^2+21\)
Vì \(\left(x+4\right)^2\ge0\left(x\in R\right)\)
nên \(-\left(x+4\right)^2\le0\left(x\in R\right)\)
do đó \(-\left(x+4\right)^2+21\le21\left(x\in R\right)\)
Vậy \(Max_E=21\)khi \(x+4=0\Leftrightarrow x=-4\)
F= \(4x-x^2+1=-\left(x^2-4x-1\right)=-\left(x^2-2.x.2+2^2-5\right)=-\left(x-2\right)^2+5\)
Vì \(\left(x-2\right)^2\ge0\left(x\in R\right)\)
nên \(-\left(x-2\right)^2\le0\left(x\in R\right)\)
do đó \(-\left(x-2\right)^2+5\le5\left(x\in R\right)\)
Vậy \(Max_F=5\)khi \(x-2=0\Leftrightarrow x=2\)
\(D=5-8x-x^2\\ =-\left[x^2+2.x.4+16\right]+21\\ =-\left(x+4\right)^2+21\le21\forall x\in R\\ \Rightarrow max_D=21.khi.x=-4\)
\(E=4x-x^2+1\\ =-\left(x^2-2.x.2+4^2\right)+17\\ =-\left(x-2\right)^2+17\le17\forall x\in R\\ Vậy:max_E=17.khi.\left(x-2\right)=0\Leftrightarrow x=2\)