tìm GTLN
M=-4x^2 + 6x -9
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13 tháng 7 2019
\(P=x^4-6x^3+10x^2-6x+9\)
\(P=\left(x^4-6x^3+9x^2\right)+\left(x^2-6x+9\right)\)
\(P=x^2\left(x^2-6x+9\right)+\left(x^2-6x+9\right)=\left(x^2+1\right)\left(x-3\right)^2\ge0\)Dấu "=" xảy ra khi x=3
\(M=\frac{3}{4x^2-4x+5}=\frac{3}{4x^2-4x+1+4}=\frac{3}{\left(2x-1\right)^2+4}\le\frac{3}{4}\)
Dấu "=" xảy ra khi x=\(\frac{1}{2}\)
\(A=\frac{2}{6x-5-9x^2}\Rightarrow-A=\frac{2}{9x^2-6x+5}=\frac{2}{9x^2+6x+1+4}=\frac{2}{\left(3x+1\right)^2+4}\le\frac{1}{2}\Rightarrow A\ge-\frac{1}{2}\)Dấu "=" xảy ra khi \(x=-\frac{1}{3}\)
TD
1
AH
Akai Haruma
Giáo viên
21 tháng 7 2021
Lời giải:
$C=-15-x^2+6x=-6-(x^2-6x+9)=-6-(x-3)^2$
Vì $(x-2)^2\geq 0$ với mọi $x\in\mathbb{R}$
$\Rightarrow C\leq -6< 0$
Vậy $C$ luôn âm.
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
Y
23 tháng 8 2018
Bài 1:
\(A=-x^2-2x+9\)
\(A=-\left(x^2+2x-9\right)\)
\(A=-\left(x^2+2x+1-10\right)\)
\(A=-\left(x+1\right)^2+10\)
Vì \(-\left(x+1\right)^2\le0\) với mọi x
\(\Rightarrow-\left(x+1\right)^2+10\le10\)
\(\Rightarrow Amax=10\Leftrightarrow x=-1\)
\(B=-9x^2+6x+25\)
\(B=-\left(9x^2-6x-25\right)\)
\(B=-\left[\left(3x\right)^2-2.3x+1-26\right]\)
\(B=-\left(3x-1\right)^2+26\)
Vì \(-\left(3x-1\right)^2\le0\) với mọi x
\(\Rightarrow-\left(3x-1\right)^2+26\le26\)
\(\Rightarrow Bmax=26\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)
\(C=-x^2+x+1\)
\(C=-\left(x^2-x-1\right)\)
\(C=-\left(x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}-1\right)\)
\(C=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\)
Vì \(-\left(x-\dfrac{1}{2}\right)^2\le0\) với mọi x
\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)
\(\Rightarrow Cmax=\dfrac{5}{4}\Leftrightarrow x=\dfrac{1}{2}\)
\(D=-2x^2+3x+1\)
\(D=-2\left(x^2-\dfrac{3}{2}x-\dfrac{1}{2}\right)\)
\(D=-2\left(x^2-2.x\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}-\dfrac{1}{2}\right)\)
\(D=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\)
Vì \(-2\left(x-\dfrac{3}{4}\right)^2\le0\) với mọi x
\(\Rightarrow-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\le\dfrac{17}{8}\)
\(\Rightarrow Dmax=\dfrac{17}{8}\Leftrightarrow x=\dfrac{3}{4}\)
\(E=-25x^2-10x+7\)
\(E=-\left(25x^2+10x-7\right)\)
\(E=-\left[\left(5x\right)^2+2.5x+1-8\right]\)
\(E=-\left(5x+1\right)^2+8\)
Vì \(-\left(5x+1\right)^2\le0\) với mọi x
\(\Rightarrow-\left(5x+1\right)^2+8\le8\)
\(\Rightarrow Emax=8\Leftrightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)
Bài 2:
\(A=9x^2+6x+4\)
\(A=\left(3x\right)^2+2.3x+1+3\)
\(A=\left(3x+1\right)^2+3\)
Vì \(\left(3x+1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(3x+1\right)^2+3\ge3\)
\(\Rightarrow Amin=3\Leftrightarrow x=-\dfrac{1}{3}\)
\(B=4x^2+4x+12\)
\(B=\left(2x\right)^2+2.2x+1+11\)
\(B=\left(2x+1\right)^2+11\)
Vì \(\left(2x+1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(2x+1\right)^2+11\ge11\)
\(\Rightarrow Bmin=11\Leftrightarrow x=-\dfrac{1}{2}\)
\(C=x^2+x+3\)
\(C=x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+3\)
\(C=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
\(\Rightarrow Cmin=\dfrac{11}{4}\Leftrightarrow x=-\dfrac{1}{2}\)
\(D=2x^2+3x+1\)
\(D=2\left(x^2+\dfrac{3}{2}x+\dfrac{1}{2}\right)\)
\(D=2\left(x^2+2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}+\dfrac{1}{2}\right)\)
\(D=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\)
Vì \(2\left(x+\dfrac{3}{4}\right)^2\ge0\) với mọi x
\(\Rightarrow2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\ge-\dfrac{1}{8}\)
\(\Rightarrow Dmin=-\dfrac{1}{8}\Leftrightarrow x=-\dfrac{3}{4}\)
\(E=64x^2+16x+3\)
\(E=\left(8x\right)^2+2.8x+1+2\)
\(E=\left(8x+1\right)^2+2\)
Vì \(\left(8x+1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(8x+1\right)^2+2\ge2\)
\(\Rightarrow Emin=2\Leftrightarrow x=-\dfrac{1}{8}\)
14 tháng 7 2019
\(A=\left(x^2+x+1\right)^2=\left[\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}\right]^2=\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]^2\ge\frac{9}{16}\)\("="\Leftrightarrow x=-\frac{1}{2}\)
\(B=x^4-6x^3+10x^2-6x+9\)
\(B=\left(x^4-6x^3+9x^2\right)+\left(x^2-6x+9\right)\)
\(B=x^2\left(x^2-6x+9\right)+\left(x^2-6x+9\right)=\left(x^2+1\right)\left(x-3\right)^2\ge0\)\("="\Leftrightarrow x=3\)
\(M=\frac{3}{4x^2-4x+5}=\frac{3}{4x^2-4x+1+4}=\frac{3}{\left(2x-1\right)^2+4}\le\frac{3}{4}\)
\("="\Leftrightarrow x=\frac{1}{2}\)
MN
14 tháng 7 2019
bạn giải thích bài 2 hộ mình, tại sao lại có ≤ \(\frac{3}{4}\)vậy? mình đi học thấy nhiều đứa viết thế cô hỏi ở đâu ra mà ko bt.
LT
0
TT
3
8 tháng 7 2017
Ta có : M = x2 + 6x - 1
=> M = x2 + 6x + 9 - 10
=> M = (x + 3)2 - 10
Mà : (x + 3)2 \(\ge0\forall x\)
Nên M = (x + 3)2 - 10 \(\ge-10\forall x\)
Vậy Mmin = -10 , dấu "=" sảy ra khi x = -3
8 tháng 7 2017
\(M=x^2+6x-1=\left(x^2+6x+9\right)-10=\left(x+3\right)^2-10\ge-10\)
Vậy \(MinM=-10\Leftrightarrow\left(x+3\right)^2=0\Leftrightarrow x=-3\)
\(N=10y-5y^2-3=-5\left(y^2-2y+1\right)+5-3=-5\left(y-1\right)^2+2\le2\)
Vậy \(MaxN=2\Leftrightarrow-5\left(y-1\right)^2=0\Leftrightarrow y=1\)
\(P=x^2-4x+y^2-8y+6=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)
\(=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
Vậy \(MinP=-14\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}}\)
\(M=-4x^2+6x-9\)
\(=-\left(4x^2-6x+9\right)=-\left(4x^2-2.2x\frac{3}{2}+\frac{9}{4}+\frac{27}{4}\right)=-\left(2x-\frac{3}{2}\right)^2-\frac{27}{4}\)
Ta có: \(\left(2x-\frac{3}{2}\right)^2\ge0\forall x\Rightarrow-\left(2x-\frac{3}{2}\right)^2\le0\Rightarrow-\left(2x-\frac{3}{2}\right)^2-\frac{27}{4}\le\frac{-27}{4}\)
Dấu '' = '' xảy ra khi: \(2x-\frac{3}{2}=0\Rightarrow x=\frac{3}{4}\)
`M= -4x^2 +6x-9`
`->M = -(4x^2 - 6x +9) =- [(2x)^2 - 2 . 2x . 3/2 + 9/4 +27/4] = -(2x - 3/2)^2 - 27/4 =< (-27)/4`
Dấu "=" xảy ra khi : `<=> (2x-3/2)^2=0 <=>x=3/4`
Vậy `max M=(-27)/4 <=> x=3/4`