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\(A=2\left(x^2-2xy+y^2\right)+\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{8067}{4}\)
\(A=2\left(x-y\right)^2+\left(x-\dfrac{3}{2}\right)^2+\dfrac{8067}{4}\ge\dfrac{8067}{4}\)
\(A_{min}=\dfrac{8067}{4}\) khi \(x=y=\dfrac{3}{2}\)
\(A=x^2-x+1\)
\(A=x^2-2\cdot\frac{1}{2}x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1\)
\(A=\left(x-\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1\)
\(A=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}+1\)
\(A=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow GTNNx^2-x-1=\frac{3}{4}\)
với \(\left(x-\frac{1}{2}\right)^2=0;x=\frac{1}{2}\)
\(B=3x^2-2x+1\)
\(B=3\left(x^2-\frac{2}{3}x+\frac{1}{3}\right)\)
\(B=3\left(x^2-2\cdot\frac{1}{3}x+\left(\frac{1}{3}\right)^2-\frac{1}{9}+\frac{1}{3}\right)\)
\(B=3\left[\left(x-\frac{1}{3}\right)^2+\frac{2}{9}\right]\)
\(B=3\left(x-\frac{1}{3}\right)^2+\frac{2}{3}\)
có \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow3\left(x-\frac{1}{2}\right)^2+\frac{2}{3}\ge\frac{2}{3}\)
\(\Rightarrow GTNN3x^2-2x+1=\frac{2}{3}\)
với\(\left(x-\frac{1}{2}\right)^2=0;x=\frac{1}{2}\)
GTNN:
\(\Leftrightarrow x^2+2\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}+1\)
\(\Leftrightarrow x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy Min của biểu thức trên =3/4 khi x+1/2=0 => x=-1/2
GTLL:
\(\Leftrightarrow-3\left(x^2-\frac{7}{3}x-\frac{1}{3}\right)\)
\(\Leftrightarrow-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{49}{36}-\frac{1}{3}\right)\)
\(\Leftrightarrow-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{61}{36}\right)\)
\(\Leftrightarrow-3\left[\left(x-\frac{7}{6}\right)^2-\frac{61}{36}\right]\)
\(\Leftrightarrow-3\left(x-\frac{7}{6}\right)^2+\frac{61}{12}\le\frac{61}{12}\)
Vậy Max của biểu thức trên = 61/12 khi x-7/6=0 => x=7/6
nha . cảm ơn . chúc bạn học tốt
1, \(3x^2-5x+4\)
\(=3\left(x^2-\frac{5}{3}x\right)+1=3\left(x^2-2.\frac{5}{6}x+\frac{25}{36}\right)+\frac{23}{12}=3\left(x-\frac{5}{6}\right)^2+\frac{23}{12}\)
Ta có: \(3\left(x-\frac{5}{6}\right)^2\ge0\forall x\Leftrightarrow3\left(x-\frac{5}{6}\right)^2+\frac{23}{12}\ge\frac{23}{12}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-\frac{5}{6}\right)^2=0\Leftrightarrow x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)
Vậy minA = \(\frac{23}{12}\Leftrightarrow x=\frac{5}{6}\)
2, Bạn thử kiểm tra lại đề bài xem
a: =2(x^2+2x+9/2)
=2(x^2+2x+1+7/2)
=2(x+1)^2+7>=7
Dấu = xảy ra khi x=-1
b: \(=2\left(\dfrac{3}{2}x^2-2xy+y^2-\dfrac{3}{2}x+\dfrac{2007}{2}\right)\)
\(=2\left(x^2-2xy+y^2+\dfrac{1}{2}x^2-\dfrac{3}{2}x+\dfrac{2007}{2}\right)\)
\(=2\left(x-y\right)^2+x^2-3x+2007\)
\(=2\left(x-y\right)^2+\left(x-\dfrac{3}{2}\right)^2+2004.75>=2004.75\)
Dấu = xảy ra khi x=y=3/2
Ta có: 5x2+10y2-6xy-4x-2y +3= x2 -6xy +(3y)2 +4x2 +y2 -4x -2y +3
= (x - 3y)2 +(2x)2 -4x+1+ y2 -2y+1 +1
= (x-3y)2 + (2x -1)2 + (y-1)2 +1
Ta có :(x-3y)2 luôn lớn hơn hoặc bằng 0
(2x -1)2 luôn lớn hơn hoặc bằng 0
(y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 +1 >0
ta có:\(A=x^2+5y^2-4xy-2y+2x+2010\)
\(=x^2+4y^2+y^2-4xy-4y+2y+2x+1+1+2008\)
\(=\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+1+\left(y^2+2x+1\right)+2008\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y+1\right)^2+2008\)
\(=\left(x-2y+1\right)^2+\left(y+1\right)^2+2008\)
Vì: (x-2y+1)2+(y+1)>0 với \(\forall x;y\)
do đó: (x-2y+1)2+(y+1)+2008 > 2008 với \(\forall x;y\)
Dấu "=" xảy ra khi x-2y+1=0 và y+1=0
ta có:
y+1=0=>y=0-1=>y=-1
thay y=-1 và x-2y+1=0
=>x-2.(-1)+1=0
=>x+2+1=0
=>x+2=-1
=>x=-1-2
=>x=-3
vậy \(A_{min}=2008\) khi x=-3 hoặc x=-1
tìm giá trị nhỏ nhất của biểu thức
A=3x2-4xy+2y2-3x+2014
giúp mik nha mik cần gấp lắm sáng mai lộp rồi
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
Ta có A = (3x + 2)2 + (x2 + y2 - 2xy) - (2x - 2y) + 2015
= (3x + 2)2 + (x - y)2 - 2(x - y) + 1 + 2014
= (3x + 2)2 + (x - y - 1)2 + 2014 \(\ge\)2014
Dấu "=" xảy ra <=> \(\hept{\begin{cases}3x+2=0\\x-y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{2}{3}\\y=x-1\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{2}{3}\\y=-\frac{5}{3}\end{cases}}\)
Vậy Min A = 2015 <=> x = -2/3 ; y = -5/3
\(A=\left(3x+2\right)^2+x^2+y^2-2xy-2x+2y+2015\)
\(=\left(3x+2\right)^2+\left(x^2-2xy+y^2\right)-\left(2x-2y\right)+1+2014\)
\(=\left(3x+2\right)^2+\left(x-y\right)^2-2\left(x-y\right)+1+2014\)
\(=\left(3x+2\right)^2+\left(x-y-1\right)^2+2014\)
Vì \(\left(3x+2\right)^2\ge0\forall x\); \(\left(x-y-1\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(3x+2\right)^2+\left(x-y-1\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(3x+2\right)^2+\left(x-y-1\right)^2+2014\ge2014\forall x,y\)
hay \(A\ge2014\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}3x+2=0\\x-y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=-2\\y=x-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-2}{3}\\y=\frac{-5}{3}\end{cases}}\)
Vậy \(minA=2014\)\(\Leftrightarrow x=-\frac{2}{3}\)và \(y=-\frac{5}{3}\)
\(A=2\left(x^2-2xy+y^2\right)+\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{8067}{4}\)
\(A=2\left(x-y\right)^2+\left(x-\dfrac{3}{4}\right)^2+\dfrac{8067}{4}\ge\dfrac{8067}{4}\)
\(A_{min}=\dfrac{8067}{4}\) khi \(x=y=\dfrac{3}{2}\)