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\(A=y^2-2y\left(x-1\right)+\left(x-1\right)^2-\left(x-1\right)^2+2x^2+4x+5\)
\(A=\left(y-x+1\right)^2-x^2+2x-1+2x^2+4x+5\)
\(A=\left(y-x+1\right)^2+x^2+6x+9-5\)
\(A=\left(y-x+1\right)^2+\left(x+3\right)^2-5\ge-5\)
Vậy Amin là -5 \(\Leftrightarrow\left\{{}\begin{matrix}\left(y-x+1\right)^2=0\\\left(x+3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-4\end{matrix}\right.\)
ta có : A=2x2 + y2-2xy +4x+2y+5
= (x2+y2+2y+1-2x-2xy)+(x2+6x
+9)-5
= (x-y-1)2+(x+3)2-5>=-5
Vậy Min A=-5 \(\Leftrightarrow\)x=-3; y=-4
\(\left(x^2-2xy+y^2-2x+2y+1\right)+\left(x^2+6x+\frac{9}{4}\right)+\left(5-\left(1+\frac{9}{4}\right)\right)=\left(x-y-1\right)^2+\left(x+1\right)^2+\frac{25}{4}\ge\frac{25}{4}\)
đẳng thúc khi :\(\left\{\begin{matrix}x+\frac{3}{2}=0\\x-y-1=0\end{matrix}\right.\)=> x=...;y=...
\(A=2x^2+y^2-2xy+4x+2y+5\)
\(A=\left(x^2+6x+9\right)+\left(y^2-2xy-2y+x^2-2x+1\right)-5\)
\(A=\left(x^2+6x+9\right)+\left[y^2-2y\left(x-1\right)+\left(x^2-2x+1\right)\right]-5\)
\(A=\left(x^2+6x+9\right)+\left[y^2-2y\left(x-1\right)+\left(x-1\right)^2\right]-5\)
\(A=\left(x+3\right)^2+\left(y-x+1\right)^2-5\ge-5\)
Dấu "=" xảy ra khi x=-3 và y=-4
\(A=2x^2+y^2-2xy+4x+2y+5\)
=> \(A=y^2-2y\left(x-1\right)+\left(x-1\right)^2-\left(x-1\right)^2+2x^2+4x+5\)
=> \(A=\left(y-x+1\right)^2-x^2+2x-1+2x^2+4x+5\)
=> \(A=\left(y-x+1\right)^2-x^2+6x+4\)
=> \(A=\left(y-x+1\right)^2-\left(x^2-2.x.3+9\right)+13\)
=> \(A=\left(y-x+1\right)^2-\left(x-3\right)^2+13\)
Có \(\left(y-x+1\right)^2\ge0\)
\(\left(x-3\right)^2\ge0\)
=> \(\left(y-x+1\right)^2-\left(x-3\right)^2+13\ge13\)
=> \(A\ge13\)
Vậy Amin = 13 <=> \(\hept{\begin{cases}y-x+1=0\\x-3=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=3\\y=2\end{cases}}\)
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
Câu 1 :
\(E=4x^2+y^2-4x-2y+3\)
\(E=\left(2x\right)^2-2\cdot2x\cdot1+1^2+y^2-2\cdot y\cdot1+1^2+1\)
\(E=\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=1\end{cases}}\)
Câu 2 :
\(G=x^2+2y^2+2xy-2y\)
\(G=x^2+2xy+y^2+y^2-2.y\cdot1+1^2-1\)
\(G=\left(x+y\right)^2+\left(y-1\right)^2-1\ge-1\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+1=0\\y=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=1\end{cases}}}\)
a) = 9(x2 - 2.x/2.9 + 1/324) - 9/324 +5
GTNN A = 4,97
b) = (2x +y)2 + y2 + 2018
GTNN B = 2018 khi x=0;y=0
c) = -4(x2 - 2.3x/ 4.2 + 9/16) +9/16 +10
GTLN C = 169/16
d) = -(x-y)2 - (2x +1) +1 + 2016
GTLN D = 2017
(trg bn cho bài khó dữ z, làm hại cả não tui)
\(A=x^2+2y^2-2xy+4x-2y+12\)
\(A=\left(x^2-2xy+y^2\right)+y^2+4x-2y+12\)
\(A=\left[\left(x-y\right)^2+2\left(x-y\right).2+4\right]+\left(y^2+2y+1\right)+7\)
\(A=\left(x-y+2\right)^2+\left(y+1\right)^2+7\)
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 \(A_{Min}=7\Leftrightarrow\left(x;y\right)=\left(-3;-1\right)\)