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A = x2 - 2*5x + 25 -35
= ( x - 5 ) 2 - 35
Vì ( x- 5 )2 >= 0 nên Amin = -35
Bài 1:
a)\(F=x^2+26y^2-10xy+14x-76y+59\)
\(=\left(x^2-2\cdot x\cdot5y+25y^2\right)+\left(14x-70y\right)+\left(y^2-6x+9\right)+50\)
\(=[\left(x-5y\right)^2+14\left(x-5y\right)+49]+\left(y-3\right)^2+1\)
\(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\)
Để Fmin=1 thì y=3;x=8
b)\(H=m^2-4mp+5p^2+10m-22p+28\)
\(=\left(m^2-2\cdot m\cdot2p+4p^2\right)+\left(10m-20p\right)+\left(p^2-2p+1\right)+27\)
\(=[\left(m-2p\right)^2+2\cdot\left(m-2p\right)\cdot5+25]+\left(p-1\right)^2+2\)
\(=\left(m-2p+5\right)^2+\left(p-1\right)^2+2\ge2\)
Để Hmin=2 thì p=1;m=-3
1/B=\(-\left(x^2+2y^2+2xy-2y\right)\)
=\(-\left(x^2+2xy+y^2+y^2-2y+1-1\right)\)
=\(-\left[\left(x+y\right)^2+\left(y-1\right)^2\right]+1\)<=1
Bmax=1 khi x+y=0 và y-1=0=>x=-1;y=1
2/C=\(x^2+x+\frac{1}{4}+y^2+y+\frac{1}{4}+\frac{1}{2}\)
=\(\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{1}{2}\)>=\(\frac{1}{2}\)
Cmin=\(\frac{1}{2}\)khi \(x+\frac{1}{2}=0\)và \(y+\frac{1}{2}=0\)=>\(x=y=\frac{-1}{2}\)
Ta có : A = x2 + 3x + 3
=> A = x2 + 3x + \(\frac{9}{4}+\frac{3}{4}\)
\(\Rightarrow A=\left(x+\frac{3}{2}\right)^2+\frac{3}{4}\) \(\ge\frac{3}{4}\forall x\in R\)
Vậy Amin = \(\frac{3}{4}\) khi \(x=-\frac{3}{2}\)
2) \(A=2x^2+2y^2-2xy-12y+2038\)
\(\Leftrightarrow A=x^2+x^2+y^2+y^2-2xy-12y+36+2002\)
\(\Leftrightarrow A=\left(x^2-2xy+y^2\right)+\left(y^2-12y+36\right)+x^2+2002\)
\(\Leftrightarrow A=\left(x-y\right)^2+\left(y-6\right)^2+x^2+2002\)
Vậy GTNN của \(A=2002\) khi
\(\left\{{}\begin{matrix}x-y=0\\y-6=0\\x=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-6=0\\y=6\\x=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=6\\x=0\end{matrix}\right.\)
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,M=x2+2x2+4-2
=(x+2)2-2
(x+2)2-2>=-2 voi moi x thuoc R
vay GTNN cua M=-2 khi x=-2