Tìm GTLN
A = -2x² + 0,5x - 8
tìm GTNN hoặc GTLN
A=|2x+4,5|+|x-2,7|
giúp mình với
Tìm GTNN
A= 2a2+b2-2ab=10a+42
Tìm GTLN
A= -x2-y2+2x-6x+9
2) \(A=-x^2-y^2+2x-6y+9=-\left(x^2-2x+1\right)-\left(y^2+6y+9\right)+19=-\left(x-1\right)^2-\left(y+3\right)^2+19\)
\(maxA=19\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
Tìm GTLN
A=6x-x^2+3
B=2x-6y-x^2-y^2-2
Bài 3: Tìm GTLN
a, \(A=4-x^2+2x\)
b, \(B=4x-x^2\)
a)Ta có:
\(A=4-x^2+2x=-\left(x^2-2x-4\right)=-\left(x^2-2x+1+3\right)\)
\(=-\left(x^2-2x+1\right)-3=-\left(x-1\right)^2-3\le-3\forall x\)
Vậy MaxA=-3 khi x=1
b) Ta có: \(B=4x-x^2=-\left(x^2-4x\right)=-\left(x^2-4x+4-4\right)=-\left(x-2\right)^2+4\le4\forall x\)Vậy MaxB=4 khi x=2
Bài 3: Tìm GTLN
a) Ta có: \(A=4-x^2+2x\)
\(=-\left(x^2-2x-4\right)\)
\(=-\left(x^2-2x+1-5\right)\)
\(=-\left(x-1\right)^2+5\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-1\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-1\right)^2+5\le5\forall x\)
Dấu '=' xảy ra khi x-1=0
hay x=1
Vậy: GTLN của biểu thức \(A=4-x^2+2x\) là 5 khi x=1
b) Ta có: \(B=4x-x^2\)
\(=-\left(x^2-4x\right)\)
\(=-\left(x^2-4x+4-4\right)\)
\(=-\left(x-2\right)^2+4\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2+4\le4\forall x\)
Dấu '=' xảy ra khi x-2=0
hay x=2
Vậy: GTLN của biểu thức \(B=4x-x^2\) là 4 khi x=2
Tìm gTLNA=\(-x^4+2x^2+4x+2002\)
tìm gtln
a) \(B=1-\sqrt{x^2-2x+2}\)
b) \(C=1+\sqrt{4x-x^2-2}\)
a: \(B=1-\sqrt{\left(x-1\right)^2+1}\)
(x-1)^2+1>=1
=>\(\sqrt{\left(x-1\right)^2+1}>=1\)
=>\(B< =0\)
Dấu = xảy ra khi x=1
b:
ĐKXĐ: -(x+2)^2+2>=0
=>-(x+2)^2>=2
=>(x+2)^2<=2
=>\(-\sqrt{2}-2< =x< =\sqrt{2}-2\)
\(-x^2+4x-2=-\left(x^2-4x+2\right)\)
\(=-\left(x^2-4x+4-2\right)=-\left(x-2\right)^2+2< =2\)
=>\(0< =\sqrt{4x-x^2-2}< =\sqrt{2}\)
=>1<=C<=căn 2+1
\(C_{max}=\sqrt{2}+1\Leftrightarrow x=2\)
TÌM GTLN CỦA :
a. A = -5 |4 - 0,5x| - 17
b. B = 5 - 2| 1 phần 2x rồi trừ 8|
c. D = 453edfmv
GTTD cua mot so luon luon lon hon hoac bang 0
a)=>|4-0,5x|=0
=>A=-17 ( de A dat GTLN)
b) =>2|1/2.x-8|=0
=>x=16=>B=5 (de B dat GTLN)
c) de j the ?
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ròi tau làm choooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
a. Có |4 - 0,5x| \(\ge\)0 \(\forall\)x
=> -5| 4 - 0,5x| \(\le\)0\(\forall\)x
=> -5|4 - 0,5x| - 17 \(\le\)-17\(\forall\)x
=> B \(\le\)-17\(\forall\)x
B = -17 <=> |4 - 0,5x| = 0
<=> 4 - 1/2x = 0
<=> 1/2x = 4
<=> x = 8
Vậy Bmax = -17 tại x = 8
gtnn,gtlnA=4-6x-x^2
B=3x^2-6x+1
C=5x^2-2x-3
\(A=4-6x-x^2=-\left(x^2+6x-4\right)=-\left(x^2+6x+9-13\right)\)
\(=-\left[\left(x+3\right)^2-13\right]=-\left(x+3\right)^2+13\le13\)
Vậy \(A_{max}=13\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
\(B=3x^2-6x+1=\left(\sqrt{3}x\right)^2-2.\sqrt{3}x.\sqrt{3}+3-2\)
\(=\left(\sqrt{3}x-\sqrt{3}\right)^2-2\ge-2\)
Vậy \(B_{min}=-2\Leftrightarrow\sqrt{3}x-\sqrt{3}=0\Leftrightarrow x=1\)
\(C=5x^2-2x-3=\left(\sqrt{5}x\right)^2-2.\sqrt{5}x.\frac{1}{\sqrt{5}}+\frac{1}{5}-\frac{16}{5}\)
\(=\left(\sqrt{5}x-\frac{1}{\sqrt{5}}\right)^2-\frac{16}{5}\ge-\frac{16}{5}\)
Vậy \(C_{min}=-\frac{16}{5}\Leftrightarrow\sqrt{5}x-\frac{1}{\sqrt{5}}=0\Leftrightarrow\sqrt{5}x=\frac{1}{\sqrt{5}}\Leftrightarrow x=\frac{1}{5}\)
tìm x :
0,5x - 3/2x = 3\(0,5x-\frac{3}{2}x=\frac{2}{7}\)