Tìm GTLN
a , A = 4x - x ^2 +5
b, B = x - x^2
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PA
1
10 tháng 8 2021
a: Ta có: \(-x^2+4x+5\)
\(=-\left(x^2-4x-5\right)\)
\(=-\left(x^2-4x+4-9\right)\)
\(=-\left(x-2\right)^2+9\le9\forall x\)
Dấu '=' xảy ra khi x=2
b: Ta có: \(-x^2-7x+4\)
\(=-\left(x^2+7x-4\right)\)
\(=-\left(x^2+2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{65}{4}\right)\)
\(=-\left(x+\dfrac{7}{2}\right)^2+\dfrac{65}{4}\le\dfrac{65}{4}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{7}{2}\)
6 tháng 1 2021
a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)
\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)
\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)
Vậy MaxQ=10 khi x=2, y=-2
b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)
\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)
Vậy MaxA=14 khi x=-3
+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)
\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)
\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)
Vậy MaxB=5 khi x=-1/2, y=1/3
c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)
\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)
Vậy MinP=2 khi x=1, y=-3
DD
Đoàn Đức Hà
Giáo viên
26 tháng 6 2021
\(A=\frac{5x^2+4x-1}{x^2}=\frac{9x^2-\left(4x^2-4x+1\right)}{x^2}=9-\frac{\left(2x-1\right)^2}{x^2}\le9\)
Dấu \(=\)khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\).
\(B=\frac{x^2}{x^2+x+1}=\frac{3x^2}{3x^2+3x+3}=\frac{4x^2+4x+4-\left(x^2+4x+4\right)}{3x^2+3x+3}=\frac{4}{3}-\frac{\left(x+2\right)^2}{3\left(x^2+x+1\right)}\le\frac{4}{3}\)
Dấu \(=\)khi \(x+2=0\Leftrightarrow x=-2\).
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 4 2021
a.
\(A=\dfrac{2013}{x^2}-\dfrac{2}{x}+1=2013\left(\dfrac{1}{x}-\dfrac{1}{2013}\right)^2+\dfrac{2012}{2013}\ge\dfrac{2012}{2013}\)
Dấu "=" xảy ra khi \(x=2013\)
b.
\(B=\dfrac{4x^2+2-4x^2+4x-1}{4x^2+2}=1-\dfrac{\left(2x-1\right)^2}{4x^2+2}\le1\)
\(B_{max}=1\) khi \(x=\dfrac{1}{2}\)
\(B=\dfrac{-2x^2-1+2x^2+4x+2}{4x^2+2}=-\dfrac{1}{2}+\dfrac{\left(x+1\right)^2}{2x^2+1}\ge-\dfrac{1}{2}\)
\(B_{max}=-\dfrac{1}{2}\) khi \(x=-1\)
MH
1 tháng 9 2021
a)A=4(x+11/8)^2 -153/16
Min A=-153/16 khi x=-11/8
b)B=3(x-1/3)^2 -4/3
Min B=-4/3 khi x=1/3
1 tháng 9 2021
Bài 1:
a) \(A=4x^2+11x-2=\left(4x^2+11x+\dfrac{121}{16}\right)-\dfrac{153}{16}=\left(2x+\dfrac{11}{4}\right)^2-\dfrac{153}{16}\ge-\dfrac{153}{16}\)
\(minA=-\dfrac{153}{16}\Leftrightarrow x=-\dfrac{11}{8}\)
b) \(B=3x^2-2x-1=3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{4}{3}=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\)
\(minB=-\dfrac{4}{3}\Leftrightarrow x=\dfrac{1}{3}\)
Bài 2:
a) \(A=-x^2+3x-1=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{5}{4}=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)
\(maxA=\dfrac{5}{4}\Leftrightarrow x=\dfrac{3}{2}\)
b) \(B=-x^2-4x+7=-\left(x^2+4x+4\right)+11=-\left(x+2\right)^2+11\le11\)
\(maxB=11\Leftrightarrow x=-2\)
25 tháng 6 2021
`A=-x^2+2x+10`
`=-(x^2-2x)+10`
`=-(x-1)^2+11<=11`
Dấu "=" xảy ra khi `x=1`.
`B=4x-2x^2+8`
`=-2(x^2-2x)+8`
`=-2(x^2-2x+1)+10`
`=-2(x-1)^2+10<=10`
Dấu "=" xảy ra khi `x=1`
`C=-x^2-x+1`
`=-(x^2+x)+1`
`=-(x^2+x+1/4)+1+1/4`
`=-(x+1/2)^2+5/4<=5/4`
Dấu "=" xảy ra khi `x=-1/2`
`D=-4x^2+6x+3`
`=-(4x^2-6x)+3`
`=-(4x^2-6x+9/4)+21/4`
`=-(2x-3/2)^2+21/4<=21/4`
Dấu "=' xảy ra khi `2x=3/2<=>x=3/4`
25 tháng 6 2021
\(a,A=-x^2+2x+10=-x^2+2x-1+11=-\left(x^2-2x+1\right)+11\)
\(=11-\left(x-1\right)^2\)
- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)
\(\Rightarrow A=11-\left(x-1\right)^2\le11\)
Vậy MaxA = 11 <=> x = 1 .
\(b,B=-2x^2+4x-2+10=-2\left(x^2-2x+1\right)+10=10-2\left(x-1\right)^2\)
- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)
\(\Rightarrow B=10-2\left(x-1\right)^2\le10\)
Vậy MaxB = 10 <=> x = 1 .
\(c,C=-x^2-\dfrac{1}{2}.2.x-\dfrac{1}{4}+\dfrac{5}{4}=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\)
- Thấy : \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\in R\)
\(\Rightarrow C=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\le\dfrac{5}{4}\)
Vậy MaxC = 5/4 <=> x = -1/2 .
\(d,D=-4x^2+6x+3=-4x^2+2x.2.\dfrac{6}{4}-\dfrac{9}{4}+\dfrac{21}{4}=-\left(4x^2-6x+\dfrac{9}{4}\right)+\dfrac{21}{4}\)
\(=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\)
- Thấy : \(\left(2x-\dfrac{3}{2}\right)^2\ge0\forall x\in R\)
\(\Rightarrow A=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\le\dfrac{21}{4}\)
Vậy MaxD=21/4 <=> x = 3/4 .
TD
2
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
1) \(A=4x-x^2+3\)
\(A=-\left(x^2-4x-3\right)\)
\(A=-\left(x^2-4x+4\right)+7\)
\(A=-\left(x-2\right)^2+7\)
Mà: \(-\left(x-2\right)^2\le0\forall x\) nên: \(A=-\left(x-2\right)^2+7\le7\)
Dấu "=" xảy ra:
\(-\left(x-2\right)^2+7=7\)
\(\Rightarrow x=2\)
Vậy: \(A_{max}=7\) khi \(x=2\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
2) \(B=x-x^2\)
\(B=-x^2+x\)
\(B=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(B=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(-\left(x-\dfrac{1}{2}\right)^2\le0\forall x\) nên \(B=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
Dấu "=" xảy ra:
\(-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{2}\)
Vậy: \(B_{max}=\dfrac{1}{4}\) với \(x=\dfrac{1}{2}\)
A = 4x - x2 + 5
A = - (x2 - 4x + 4 ) +1
A = - ( x -2 )2 + 1
Do - (x - 2 )2 <= 0
=> A <= 1
Dấu "=" xảy ra khi x -2 =0
<=> x = 2
Vậy A min = 1 khi x =2
\(A=4x-x^2+5=-\left(x^2-4x-5\right)=-\left(x^2-4x+4-9\right)=-\left(x-2\right)^2+9\le9\forall x\)
dấu = xảy ra khi: \(-\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy Max A = 9 tại x = 2
\(B=x-x^2=-\left(x^2+x\right)=-\left(x^2+x+\frac{1}{4}-\frac{1}{4}\right)=-\left(x+\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
dấu = xảy ra khi: \(-\left(x+\frac{1}{2}\right)^2=0\Rightarrow x=-\frac{1}{2}\)
Vậy Max B = 1/4 tại x= -1/2
=.= hok tốt!!