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Bài làm:
a) Sửa đề:
\(A=4x-x^2=-\left(x^2-4x+4\right)+4\)
\(=-\left(x-2\right)^2+4\le4\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(-\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy \(A_{Max}=4\Leftrightarrow x=2\)
b) \(B=-x^2-4x+5=-\left(x^2+4x+4\right)+9\)
\(=-\left(x+2\right)^2+9\le9\)
Dấu "=" xảy ra khi: \(-\left(x+2\right)^2=0\Rightarrow x=-2\)
Vậy \(B_{Max}=9\Leftrightarrow x=-2\)
c) \(C=-x^2-2y^2-2xy+2y\)
\(C=-\left(x^2+2xy+y^2\right)-\left(y^2-2y+1\right)+1\)
\(C=-\left(x+y\right)^2-\left(y-1\right)^2+1\le1\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}-\left(x+y\right)^2=0\\-\left(y-1\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)
Vậy \(C_{Max}=1\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)
a) Sửa : A = 4x - x2
A = -x2 + 4x - 4 + 4
A = -( x2 - 4x + 4 ) + 4
A = -( x - 2 )2 + 4
-( x - 2 )2 ≤ 0 ∀ x => -( x - 2 ) + 4 ≤ 4
Dấu " = " xảy ra <=> x - 2 = 0 => x = 2
Vậy AMax = 4 , đạt được khi x = 2
b) B = -x2 - 4x + 5 = -x2 - 4x - 4 + 9 = -( x2 + 4x + 4 ) + 9 = -( x + 2 )2 + 9
-( x + 2 )2 ≤ 0 ∀ x => -( x + 2 )2 + 9 ≤ 9
Dấu " = " xảy ra <=> x + 2 = 0 => x = -2
Vậy BMax = 9, đạt được khi x = -2
c) C = -x2 - 2y2 - 2xy + 2y
= ( -x2 - 2xy - y2 ) + ( -y2 + 2y -1 ) + 1
= -( x2 + 2xy + y2 ) - ( y2 - 2y + 1 ) + 1
= -( x + y )2 - ( y - 1 )2 + 1
\(\hept{\begin{cases}-\left(x+y\right)^2\le0\\-\left(y-1\right)^2\le0\end{cases}\Rightarrow}-\left(x+y\right)^2-\left(y-1\right)^2+1\le1\forall x,y\)
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+y=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x+y=0\\y=1\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)
Vậy CMax = 1 , đạt được khi x = -1 ; y = 1
a) Đặt \(A=x^2+4x+7\)
\(A=\left(x^2+4x+4\right)+3\)
\(A=\left(x+2\right)^2+3\)
Mà \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow A\ge3>0\)
b) Đặt \(B=4x^2-4x+5\)
\(B=\left(4x^2-4x+1\right)+4\)
\(B=\left(2x-1\right)^2+4\)
Mà \(\left(2x-1\right)^2\ge0\forall x\)
\(\Rightarrow B\ge4>0\)
c) Đặt \(C=x^2+2y^2+2xy-2y+3\)
\(C=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+2\)
\(C=\left(x+y\right)^2+\left(y-1\right)^2+2\)
Mà \(\left(x+y\right)^2\ge0\forall x;y\)
\(\left(y-1\right)^2\ge0\forall y\)
\(\Rightarrow C\ge2>0\)
\(A=2x^2+y^2-2xy-2x+3\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+2\)
\(A=\left(x-y\right)^2+\left(x-1\right)^2+2\)
Mà \(\left(x-y\right)^2\ge0\forall x;y\)
\(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow A\ge2\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}x-y=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=1\\x=1\end{cases}}\)
Vậy Min A = 2 khi x=y=1
\(1,a,A=x^2-6x+25\)
\(=x^2-2.x.3+9-9+25\)
\(=\left(x-3\right)^2+16\)
Ta có :
\(\left(x-3\right)^2\ge0\)Với mọi x
\(\Rightarrow\left(x-3\right)^2+16\ge16\)
Hay \(A\ge16\)
\(\Rightarrow A_{min}=16\)
\(\Leftrightarrow x=3\)
\(A=x^2+2y^2-2xy+4x-2y+12\)
\(A=\left(x^2-2xy+y^2\right)+y^2+4x-2y+12\)
\(A=\left[\left(x-y\right)^2+2\left(x-y\right).2+4\right]+\left(y^2+2y+1\right)+7\)
\(A=\left(x-y+2\right)^2+\left(y+1\right)^2+7\)
Mà \(\left(x-y+2\right)^2\ge0\forall x;y\)
\(\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow A\ge7\)
Dấu "=" xảy ra khi : \(\hept{\begin{cases}x-y+2=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=-1\end{cases}}\)
Vậy \(A_{Min}=7\Leftrightarrow\left(x;y\right)=\left(-3;-1\right)\)
Câu 1:
a: \(C=a^2+b^2=\left(a+b\right)^2-2ab=23^2-2\cdot132=265\)
b: \(D=x^3+y^3+3xy\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\)
\(=1-3xy+3xy=1\)
<=> x^2 + 2x(y+2) + y^2+4y+4+y^2+2y+1-4
<=> x^2 + 2x(y+2) + (y+2)^2 + (y+1)^2 - 4
<=> (x+y+2)^2 + (y+1)^2 - 4 >= -4
min = -4 khi y = -1 , x = -1
\(=\left(x+y+2\right)^2+\left(y+1\right)^2-4\)
Vì \(\left(x+y+2\right)^2\ge0\forall x\) , \(\left(y+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+y+2\right)^2+\left(y+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+y+2\right)^2+\left(y+1\right)^2-4\ge-4\forall x\)
Vậy GTNN của A=-4 Dấu bằng xảy ra khi
\(\Rightarrow\hept{\begin{cases}\left(x+y+2\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2-y\\y=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=-1\end{cases}}\)
Vậy GTNN của A=-4 khi và chỉ khi x=-3 , y=-1
a/ \(A=4x^2+y^2-4x-2y+3\)
\(=\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)+1\)
\(=\left(2x-1\right)^2+\left(y-1\right)^2+1\)
Với mọi x, y ta có :
\(\left\{{}\begin{matrix}\left(2x-1\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\)
\(\Leftrightarrow A\ge1\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-1\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=1\end{matrix}\right.\)
Vậy...
b/ \(B=x^2+2y^2+2xy-2y\)
\(=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)-1\)
\(=\left(x+y\right)^2+\left(y-1\right)^2-1\)
Với mọi x, y ta có :
\(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2-1\ge0\)
\(\Leftrightarrow B\ge-1\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Vậy..
a.\(A=4x^2+y^2-4x-2y+3\)
\(A=\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)+1\)
\(A=\left(2x-1\right)^2+\left(y-1\right)^2+1\)
Vì \(\left(2x-1\right)^2\ge0\) và \(\left(y-1\right)^2\ge0\)
\(\Rightarrow\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\)
\(\Rightarrow Min_A=1\) khi \(\left\{{}\begin{matrix}2x-1=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=1\end{matrix}\right.\)
b.\(B=x^2+2y^2+2xy-2y\)
\(B=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)-1\)
\(B=\left(x+y\right)^2+\left(y-1\right)^2-1\)
Vì \(\left(x+y\right)^2\ge0\) và \(\left(y-1\right)^2\ge0\)
\(\Rightarrow\left(x+y\right)^2+\left(y-1\right)^2-1\ge-1\)
\(\Rightarrow Min_B=-1\) khi \(\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-y\\y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Vậy \(Min_B=-1\) khi \(x=-1;y=1\)