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Ta có: \(B=x^2-4xy+5y^2-22y+28\)
\(=x^2-4xy+y^2-22y+121-93\)
\(=\left(x-2y\right)^2+\left(y-11\right)^2-93\)
Vì \(\left(x-2y\right)^2\ge0;\left(y-11\right)^2\ge0\)
\(\Rightarrow B\ge-93\)
Dấu "=" xảy ra khi \(y-11=0\Rightarrow y=11\)
\(x-2y=0\Rightarrow x-2.11=0\Rightarrow x=22\)
Vậy Bmin=-93 khi x=22; y=11
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\(D=x^2+5y^2-2xy+4y+3\)
\(=x^2-2xy+y^2+4y^2+4y+1+2\)
\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2\)
\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(2y+1\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(2y+1\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{1}{2}\)
Vậy \(D_{min}=2\Leftrightarrow x=y=-\dfrac{1}{2}\)
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d= x2 + 5y2 + 2xy - 2y + 2005
d= x2 + 2xy + y2 + 4y2 - 2y + \(\frac{1}{4}+\)
d= ( x+ y )2 + ( 2y - \(\frac{1}{2}\))2 + \(\frac{8019}{4}\)\(\ge\)\(\frac{8019}{4}\)
dmin= \(\frac{8019}{4}khi\hept{\begin{cases}y=\frac{1}{4}\\x=-y=\frac{-1}{4}\end{cases}}\)
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Đặt `A=2x^2+2xy+5y^2-8x-22y`
`<=>2A=4x^2+4xy+10y^2-16x-44y`
`<=>2A=4x^2+4xy+y^2-8(2x+y)+9y^2-28y`
`<=>2A=(2x+y)^2-8(2x+y)+16+9y^2-28y+196/9-196/9`
`<=>2A=(2x+y-4)^2+(3y-14/3)^2-196/9>=-196/9`
`<=>A>=-98/9`
Dấu "=" xảy ra khi `y=14/9,x=(4-y)/2=11/9`
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\(B=2x^2+2xy+5y^2-8x-22y\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+16\right)+\left(4y^2-22y+\frac{484}{16}\right)-\frac{185}{4}\)
\(=\left(x+y\right)^2+\left(x-4\right)^2+\left(2y-\frac{22}{4}\right)^2-\frac{185}{4}\ge-\frac{185}{4}\)
Dấu = xảy ra khi :
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Bn tự giải nốt nhé, mk ko bt có đúng hay ko , nếu sai thì thông cảm nha........
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Đặt A bằng biểu thức trên.
Ta có: A = 2x2 + x(2y - 8) + (5y2 - 22y + 1)
\(\Leftrightarrow2x^2+2x\left(y-4\right)+\left(5y^2-22y+1-A\right)=0\)
+) Nếu x = 0: Khi đó \(A=5y^2-22y+1=5\left(y-\frac{11}{5}\right)^2-\frac{116}{5}\ge-\frac{116}{5}\).
+) Nếu x \(\ne\) 0: Xét pt bậc 2 đối với x. Để pt có nghiệm thì:
\(\Delta'=(y-4)^2-2(5y^2-22y+1-A)\geq0\)
\(\Leftrightarrow2A\geq9y^2-36y-14=(3y-6)^2-50\geq-50\).
\(\Leftrightarrow A\ge-25\)
So sánh 2 TH, ta được min A = \(=-25\) khi và chỉ khi \(x=1;y=2\).