Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=x^2+4x+5=\left(x+2\right)^2+1\ge1\)
Dấu \("="\Leftrightarrow x=-2\)
\(B=x^2+10x-1=\left(x+5\right)^2-26\ge-26\)
Dấu \("="\Leftrightarrow x=-5\)
\(C=5-4x+4x^2=\left(2x-1\right)^2+4\ge4\)
Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)
\(D=x^2+y^2-2x+6y-3=\left(x-1\right)^2+\left(y+3\right)^2-13\ge-13\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
\(E=2x^2+y^2+2xy+2x+3=\left(x+y\right)^2+\left(x+1\right)^2+2\ge2\)
Dấu \("="\Leftrightarrow x=-y=-1\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
\(A=x^2+4x+5\)
\(=x^2+4x+4+1\)
\(=\left(x+2\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=-2
\(C=4x^2-4x+5\)
\(=4x^2-4x+1+4\)
\(=\left(2x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
Lời giải:
$M=(x^2+y^2+2xy)+x^2+y^2-6x-6y+11$
$=(x+y)^2+x^2+y^2-6x-6y+11$
$=(x+y)^2-4(x+y)+4+(x^2-2x+1)+(y^2-2y+1)+5$
$=(x+y-2)^2+(x-1)^2+(y-1)^2+5\geq 0+0+0+5=5$
Vậy $M_{\min}=5$. Giá trị này đạt tại $x+y-2=x-1=y-1=0$
$\Leftrightarrow x=y=1$
a) \(A=4x^2-12x+100=\left(2x\right)^2-12x+3^2+91=\left(2x-3\right)^2+91\)
Ta có: \(\left(2x-3\right)^2\ge0\forall x\inℤ\)
\(\Rightarrow\left(2x-3\right)^2+91\ge91\)
hay A \(\ge91\)
Dấu "=" xảy ra <=> \(\left(2x-3\right)^2=0\)
<=> 2x-3=0
<=> 2x=3
<=> \(x=\frac{3}{2}\)
Vậy Min A=91 đạt được khi \(x=\frac{3}{2}\)
b) \(B=-x^2-x+1=-\left(x^2+x-1\right)=-\left(x^2+x+\frac{1}{4}-\frac{5}{4}\right)=-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\)
Ta có: \(-\left(x+\frac{1}{2}\right)^2\le0\forall x\)
\(\Rightarrow-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\) hay B\(\le\frac{5}{4}\)
Dấu "=" \(\Leftrightarrow-\left(x+\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x+\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy Max B=\(\frac{5}{4}\)đạt được khi \(x=\frac{-1}{2}\)
\(C=2x^2+2xy+y^2-2x+2y+2\)
\(C=x^2+2x\left(y-1\right)+\left(y-1\right)^2+x^2+1\)
\(\Leftrightarrow C=\left(x+y-1\right)^2+x^2+1\)
Ta có:
\(\hept{\begin{cases}\left(x+y-1\right)^2\ge0\forall x;y\inℤ\\x^2\ge0\forall x\inℤ\end{cases}}\)
\(\Leftrightarrow\left(x+y-1\right)^2+x^2+1\ge1\)
hay C\(\ge\)1
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x+y-1\right)^2=0\\x^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=1\\x=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=1\\x=0\end{cases}}}\)
Vậy Min C=1 đạt được khi y=1 và x=0
\(D=x^2+4y^2-2xy-6y-10x+10y+32\)
\(=x^2-2.x\left(y+5\right)+\left(y+5\right)^2-\left(y+5\right)^2+4y^2+4y+32\)
\(=\left(x-y-5\right)^2-y^2-10y-25+4y^2+4y+32\)
\(=\left(x-y-5\right)^2+3y^2-6y+7\)
\(=\left(x-y-5\right)^2+3\left(y^2-2y+1\right)+4\)
\(=\left(x-y-5\right)^2+3\left(y-1\right)^2+4\)
Ta thấy : \(\left(x-y-5\right)^2+3\left(y-1\right)^2\ge0\forall x,y\)
\(\Rightarrow D\ge4\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y-5=0\\y-1=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=6\\y=1\end{cases}}\)
Vậy : min \(D=4\) tại \(x=6,y=1\)
a/ \(M=x^2+y^2-x+6y+10=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+10-\frac{1}{4}-9\)
\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Suy ra Min M = 3/4 <=> (x;y) = (1/2;-3)
b/
1/ \(A=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
Suy ra Min A = 7 <=> x = 2
2/ \(B=x-x^2=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Suy ra Min B = 1/4 <=> x = 1/2
3/ \(N=2x-2x^2-5=-2\left(x^2-x+\frac{1}{4}\right)-5+\frac{1}{2}=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\)
\(\ge-\frac{9}{2}\)
Suy ra Min N = -9/2 <=> x = 1/2
b: Tham khảo:
a: \(P=x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}=\left(x-\dfrac{5}{2}\right)^2-\dfrac{25}{4}\ge-\dfrac{25}{4}\forall x\)
Dấu '=' xảy ra khi x=5/2
Bài làm:
a) \(P=x^2-5x=\left(x^2-5x+\frac{25}{4}\right)-\frac{25}{4}\)
\(=\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\le-\frac{25}{4}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(x=\frac{5}{2}\)
Vậy \(Min_P=-\frac{25}{4}\Leftrightarrow x=\frac{5}{2}\)
a) P = x2 - 5x
= ( x2 - 5x + 25/4 ) - 25/4
= ( x - 5/2 )2 - 25/4
( x - 5/2 )2 ≥ 0 ∀ x => ( x - 5/2 )2 - 25/4 ≥ -25/4
Đẳng thức xảy ra <=> x - 5/2 = 0 => x = 5/2
=> MinF = -25/4 <=> x = 5/2
b) Q = x2 + 2y2 + 2xy - 2x - 6y + 2015
= ( x2 + 2xy + y2 - 2x - 2y + 1 ) + ( y2 - 4y + 4 ) + 2010
= [ ( x + y )2 - 2( x + y ) + 12 ] + ( y - 2 )2 + 2010
= ( x + y - 1 )2 + ( y - 2 )2 + 2010
\(\hept{\begin{cases}\left(x+y-1\right)^2\ge0\forall x,y\\\left(y-2\right)^2\ge0\forall x\end{cases}}\Rightarrow\left(x+y-1\right)^2+\left(y-2\right)^2+2010\ge2010\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x+y-1=0\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x+y-1=0\\y=2\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}\)
=> MinQ = 2010 <=> x = -1 , y = 2
\(A=\left(x^2+2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{5}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\\ A_{min}=-\dfrac{5}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ B=\left(x^2+2xy+y^2\right)+\left(x^2+6x+9\right)+3\\ B=\left(x+y\right)^2+\left(x+3\right)^2+3\ge3\\ B_{min}=3\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\\ C=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\\ C_{max}=1\Leftrightarrow x=1\)
Lời giải:
$A=(x^2-2xy+y^2)+y^2+2x-6y+2028$
$=(x-y)^2+2(x-y)+(y^2-4y)+2028$
$=(x-y)^2+2(x-y)+1+(y^2-4y+4)+2023$
$=(x-y+1)^2+(y-2)^2+2023\geq 0+0+2023=2023$
Vậy $A_{\min}=2023$.
Giá trị này đạt tại $x-y+1=y-2=0$
$\Leftrightarrow y=2; x=1$
A=(x2−2xy+y2)+y2+2x−6y+2028
=(�−�)2+2(�−�)+(�2−4�)+2028=(x−y)2+2(x−y)+(y2−4y)+2028
=(�−�)2+2(�−�)+1+(�2−4�+4)+2023=(x−y)2+2(x−y)+1+(y2−4y+4)+2023
=(�−�+1)2+(�−2)2+2023≥0+0+2023=2023=(x−y+1)2+(y−2)2+2023≥0+0+2023=2023
Vậy �min=2023Amin=2023.
Giá trị này đạt tại �−�+1=�−2=0x−y+1=y−2=0
⇔�=2;�=1⇔y=2;x=1