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Bài 5:
a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)
\(minA=5\Leftrightarrow x=2\)
b) \(B=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)
c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)
Bài 4:
a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
\(maxM=7\Leftrightarrow x=2\)
b) \(N=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)
\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)
a: \(A=-4x^2+4x-1\)
\(=-\left(4x^2-4x+1\right)\)
\(=-\left(2x-1\right)^2\le0\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
b: \(B=-x^2+5x\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}\right)+\dfrac{25}{4}\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)
\(A=x^2-4x+1\)
\(A=x^2-4x+4-3\)
\(A=\left(x-2\right)^2-3\)
Min A = -3
Min A xảy ra khi (x-2)2=0
x-2=0
x=2
A đến C là tìm GTNN
\(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)
Dấu "=" xảy ra ⇔ x=2
\(B=2x^2-x+1=2\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)+\dfrac{7}{8}=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{4}\)
\(C=x^2-x+1=\left(x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)
\(A=\dfrac{4x+3}{x^2+1}\Leftrightarrow Ax^2+A=4x+3\\ \Leftrightarrow Ax^2-4x+A-3=0\)
Coi đây là PT bậc 2 ẩn x thì PT có nghiệm
\(\Leftrightarrow\Delta=16-4A\left(A-3\right)\ge0\\ \Leftrightarrow16-4A^2+12A\ge0\\ \Leftrightarrow-A^2+3A+4\ge0\\ \Leftrightarrow-1\le A\le4\)
Vậy \(A_{max}=4;A_{min}=-1\)
\(A_{max}=4\Leftrightarrow\dfrac{4x+3}{x^2+1}=4\Leftrightarrow4x^2-4x+1=0\\ \Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\\ A_{min}=-1\Leftrightarrow\dfrac{4x+3}{x^2+1}=-1\Leftrightarrow x^2+1=-4x-3\Leftrightarrow x^2+4x+4=0\\ \Leftrightarrow\left(x+2\right)^2=0\Leftrightarrow x=-2\)
a: =-x^2+6x-4
=-(x^2-6x+4)
=-(x^2-6x+9-5)
=-(x-3)^2+5<=5
Dấu = xảy ra khi x=3
b: =3(x^2-5/3x+7/3)
=3(x^2-2*x*5/6+25/36+59/36)
=3(x-5/6)^2+59/12>=59/12
Dấu = xảy ra khi x=5/6
c: \(=-\left(x-3\right)^2+2\left|x-3\right|\)
\(=-\left[\left(\left|x-3\right|\right)^2-2\left|x-3\right|+1-1\right]\)
\(=-\left(\left|x-3\right|-1\right)^2+1< =1\)
Dấu = xảy ra khi x=4 hoặc x=2
a) \(2x^2-x+1=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{4}\)
b) \(5x-x^2+4=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{41}{4}\le\dfrac{41}{4}\)
\(ĐTXR\Leftrightarrow x=\dfrac{5}{2}\)
c) \(x^2+5y^2-2xy+4y+3=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)
\(ĐTXR\Leftrightarrow\)\(x=y=-\dfrac{1}{2}\)
b: ta có: \(-x^2+5x+4\)
\(=-\left(x^2-5x-4\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{41}{4}\right)\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{41}{4}\le\dfrac{41}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)
1: A=(x-1)^2>=0
Dấu = xảy ra khi x=1
5: B=-(x^2+6x+10)
=-(x^2+6x+9+1)
=-(x+3)^2-1<=-1
Dấu = xảy ra khi x=-3
2: B=x^2+4x+4-9
=(x+2)^2-9>=-9
Dấu = xảy ra khi x=-2
6: =-(x^2-5x-3)
=-(x^2-5x+25/4-37/4)
=-(x-5/2)^2+37/4<=37/4
Dấu = xảy ra khi x=5/2
3: =x^2+x+1/4-1/4
=(x+1/2)^2-1/4>=-1/4
Dấu = xảy ra khi x=-1/2
7: =4x^2+4x+1-2
=(2x+1)^2-2>=-2
Dấu = xảy ra khi x=-1/2
A=x2-4x+1
=x2-4x+4-3
=(x-2)2-3
=> GTNN là : -3
B=đề sai.
C=chả hiểu gì hết.