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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 \(C_{min}=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{5}{2}\)
\(A=5-8x+x^2=-8x+x^2+6-11\)
\(=\left(x-4\right)^2-11\)
Vì \(\left(x-4\right)^2\ge0\forall x\)\(\Rightarrow\left(x-4\right)^2-11\ge-11\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)
Vậy Amin = - 11 <=> x = 4
\(B=\left(2-x\right)\left(x+4\right)=-x^2-2x+8\)
\(=-\left(x^2+2x+1\right)+9=-\left(x+1\right)^2+9\)
Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+1\right)^2+9\le9\)
Dấu "=" xảy ra \(\Leftrightarrow-\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy Bmax = 9 <=> x = - 1
Ta có: x^2>=0 với mọi x =>-x^2<=0 với mọi x =>-x^2-8*5<=-40
Dấu bằng xảy ra khi x^2=0 =>x=0
`A=x^2-4x+1`
`=x^2-4x+4-3`
`=(x-2)^2-3>=-3`
Dấu "=" xảy ra khi x=2
`B=4x^2+4x+11`
`=4x^2+4x+1+10`
`=(2x+1)^2+10>=10`
Dấu "=" xảy ra khi `x=-1/2`
`C=(x-1)(x+3)(x+2)(x+6)`
`=[(x-1)(x+6)][(x+3)(x+2)]`
`=(x^2+5x-6)(x^2+5x+6)`
`=(x^2+5x)^2-36>=-36`
Dấu "=" xảy ra khi `x=0\or\x=-5`
`D=5-8x-x^2`
`=21-16-8x-x^2`
`=21-(x^2+8x+16)`
`=21-(x+4)^2<=21`
Dấu "=" xảy ra khi `x=-4`
`E=4x-x^2+1`
`=5-4+4-x^2`
`=5-(x^2-4x+4)`
`=5-(x-2)^2<=5`
Dấu "=" xảy ra khi `x=5`
\(B1,a,A=x^2-6x+11\)
\(=\left(x^2-6x+9\right)+2\)
\(=\left(x-3\right)^2+2\ge2\)
Dấu "=" <=> x=3
Vậy ..........
\(b,B=x^2-20x+101\)
\(=\left(x^2-20x+100\right)+1\)
\(=\left(x-10\right)^2+1\ge1\)
Dấu "=" <=> x = 10
Vậy .
\(2,a,A=4x-x^2+3\)
\(=7-\left(x^2-4x+4\right)\)'
\(=7-\left(x-2\right)^2\le7\)
Dấu ''='' <=> x = 2
Vậy .
\(b,B=-x^2+6x-11\)
\(=-2-\left(x^2-6x+9\right)\)
\(=-2-\left(x-3\right)^2\le-2\)
Dấu ""=" <=> x = 3
Vậy..
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.....
Nhớ cho 5 sao luôn nhé
Ta có: \(4x^2-8x+7=4x^2-8x+4+3\left(2x-2\right)^2+3\ge3\)
\(\Rightarrow B>0\)
Vậy B có GTLN \(\Leftrightarrow\left(2x-2\right)^2+3\)có GTNN
Mà \(\left(2x-2\right)^2+3\ge3\Rightarrow Min\left(4x^2=8x+7\right)=3\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
\(\Rightarrow\)Max B = 3\(\Leftrightarrow x=1\)
\(B=x^2+8x+16-16\)
\(B=\left(x+4\right)^2-16\)
có : \(\left(x+4\right)^2\ge0\Rightarrow\left(x+4\right)^2-16\ge-16\)
\(\Rightarrow B\ge-16\)
Dấu "=" xảy ra khi
(x + 4)2 = 0 => x + 4 = 0 => x = - 4
vậy Min B = -16 khi x = -4
\(B=x^2+8x\)
\(=x^2.2.x.4+16-16\)
\(=\left(x+4\right)^2-16\)
Vì \(\left(x+4\right)^2\ge0;\forall x\)
\(\Rightarrow\left(x+4\right)^2-16\ge0-16;\forall x\)
Hay\(B\ge-16;\forall x\)
Dấu "=" xảy ra\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy MIN B= -16 \(\Leftrightarrow x=-4\)
A= 8x - x2 - 11=-(x2-8x+11)=-(x2-2.4.x+16-5)=-[(x-4)2-5]=5-(x-4)2
Vì (x-4)2\(\ge0\forall x\)=>5-(x-4)2\(\le5\forall x\)
<=>maxA = 5 khi (x-4)2=0 <=> x-4=0 <=> x=4
cảm ơn bn