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\(B=4x^2-4x+7\)
\(B=\left[\left(2x\right)^2-2.2x.1+1^2\right]+6\)
\(B=\left(2x-1\right)^2+6\)
Ta có : \(\left(2x-1\right)^2\ge0\)
\(\Rightarrow\left(2x-1\right)^2+6\ge6\)
Vậy GTNN của B là \(6\)
Khi \(2x-1=0\)
\(2x=1\)
\(x=\frac{1}{2}\)
C=2x2+4x+5
C=2.(x2+2x+2,5)
C=2.(x2+2x+1)+3
C=2.(x+1)2+3
Vì 2.(x+1)2\(\ge\)0
Suy ra:2.(x+1)2+3 \(\ge\)3
Dấu = xảy ra khi x+1=0
x=-1
Vậy Min C=3 khi x=-1
bài 1
a, \(A=\frac{1}{-x^2+2x-2}=\frac{1}{-\left(x^2-2x+1\right)-1}=\frac{1}{-\left(x-1\right)^2-1}\)
Vì \(-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2-1\le-1\Rightarrow A=\frac{1}{-\left(x-1\right)^2-1}\ge\frac{1}{-1}=-1\)
Dấu "=" xảy ra khi x=1
Vậy Amin=-1 khi x=1
b, \(B=\frac{2}{-4x^2+8x-5}=\frac{2}{-4\left(x^2-2x+1\right)-1}=\frac{2}{-4\left(x-1\right)^2-1}\ge\frac{2}{-1}=-2\)
Dấu "=" xảy ra khi x=1
Vậy Bmin=-2 khi x=1
bài 2:
a, \(A=\frac{3}{2x^2+2x+3}=\frac{3}{2\left(x^2+x+\frac{1}{4}\right)+\frac{5}{2}}=\frac{3}{2\left(x+\frac{1}{2}\right)^2+\frac{5}{2}}\)
Vì \(2\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow2\left(x+\frac{1}{2}\right)^2+\frac{5}{2}\ge\frac{5}{2}\Rightarrow A=\frac{3}{2\left(x+\frac{1}{2}\right)^2+\frac{5}{2}}\le\frac{3}{\frac{5}{2}}=\frac{6}{5}\)
dấu "=" xảy ra khi x=-1/2
Vậy Amax=6/5 khi x=-1/2
b, \(B=\frac{5}{3x^2+4x+15}=\frac{5}{3\left(x^2+\frac{4}{3}x+\frac{4}{9}\right)+\frac{41}{3}}=\frac{5}{3\left(x+\frac{2}{3}\right)^2+\frac{41}{3}}\le\frac{5}{\frac{41}{3}}=\frac{15}{41}\)
Dấu '=" xảy ra khi x=-2/3
Vậy Bmax=15/41 khi x=-2/3
A=x2-4x+7
= x2-4x+4+3
= (x-2)2+3
Vì (x+2)2>/ 0
Nên (x-2)2+3>/3
Vậy MAX của A=3 khi x-2=0 => x=2
a) \(A=5x^2-4x+1\)
\(=5\left(x^2-\frac{4}{5}x+\frac{1}{5}\right)\)
\(=5\left(x^2-\frac{4}{5}x+\frac{4}{25}-\frac{2}{25}\right)\)
\(=5\left[\left(x-\frac{2}{5}\right)^2-\frac{2}{25}\right]\)
\(=5\left[\left(x-\frac{2}{5}\right)^2\right]-2\ge-2\)
Vậy \(A_{min}=-2\Leftrightarrow x-\frac{2}{5}=0\Leftrightarrow x=\frac{2}{5}\)
Sửa)):Dòng 3
\(=5\left(x^2-\frac{4}{5}x+\frac{4}{25}+\frac{1}{25}\right)\)
\(=5\left[\left(x-\frac{2}{5}\right)^2+\frac{1}{25}\right]\)
\(=5\left[\left(x-\frac{2}{5}\right)^2\right]+\frac{1}{5}\ge\frac{1}{5}\)
(Dấu "="\(\Leftrightarrow x-\frac{2}{5}=0\Leftrightarrow x=\frac{2}{5}\)
a, (x-1)(x-3)+11
=x2-3x-x+3+11
=(x-2)2+10
Vì..................................
b,5-4x2+4x
=-(4x2-4x+4)+9
=-(2x-2)2+9
...........................................................
Bài làm:
Ta có: \(4x^2+2y^2+4xy-4x-8y+15\)
\(=\left(4x^2+4xy+y^2\right)-2\left(2x+y\right)+1+y^2-6y+9+5\)
\(=\left(2x+y\right)^2-2\left(2x+y\right)+1+\left(y-3\right)^2+5\)
\(=\left(2x+y-1\right)^2+\left(y-3\right)^2+5\ge5\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(2x+y-1\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=3\end{cases}}\)
Vậy \(Min=5\Leftrightarrow\hept{\begin{cases}x=-1\\y=3\end{cases}}\)
4x2 + 2y2 + 4xy - 4x - 8y + 15
= [ ( 4x2 + 4xy + y2 ) - 2( 2x + y ) + 1 ] + ( y2 - 6y + 9 ) + 5
= ( 2x + y - 1 )2 + ( y - 3 )2 + 5
\(\hept{\begin{cases}\left(2x+y-1\right)^2\ge0\forall x,y\\\left(y-3\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(2x+y-1\right)^2+\left(y-3\right)^2+5\ge5\forall x,y\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}2x+y-1=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=3\end{cases}}\)
Vậy GTNN của biểu thức = 5 <=> x = -1 ; y = 3
\(C=-\left(4x^2-4x-15\right)\)
\(=-\left(4x^2-4x+1-16\right)\)
\(=-\left(2x-1\right)^2+16< =16\)
Dấu = xảy ra khi x=1/2
\(D=x^2-4x+3+21\)
\(=x^2-4x+4+20=\left(x-2\right)^2+20>=20\)
Dấu '=' xảy ra khi x=2
2x2+4x=2(x2+2x+1)−2=2(x+1)2−12x2+4x=2(x2+2x+1)−2=2(x+1)2−1
2(x2+1)≥0⇒2(x2+1)−2≥−22(x2+1)≥0⇒2(x2+1)−2≥−2⇒min=−2⇔x+1=0⇒x=−1
Ta có:
\(2x^2+4x+15\)
\(=2\left(x^2+2x+1\right)+13\)
\(=2\left(x+1\right)^2+13\)
\(\ge13\) Dấu "=" xảy ra tại \(x=-1\)
BÙI THỊ YẾN NHI Copy mak cx stupid
Ta có : \(-4x^2+4x-5=-\left(4x^2-4x+5\right)=-\left(2x-1\right)^2-4\le-4\)
\(\Rightarrow B\ge\dfrac{15}{-4}\)
Dấu ''='' xảy ra khi x = 1/2
Vậy GTNN B là -15/4 khi x = 1/2