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x^2-x+1+x^2-x-2=2x^2-2x+1=1-2x+x^2 + x^2= (1-x)^2 + x^2
Ta có (1-x)^2 +x^2>=x^2
Khi 1-x=0 <=> x=1
Vậy GTNN là 1 khi 1-x=0 <=> x=1
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Bài 1:
\(x^2-x+1=x^2-x+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" khi \(x=\frac{1}{2}\)
Vậy \(Min=\frac{3}{4}\) khi \(x=\frac{1}{2}\)
Bài 2:
\(x^2+10x+2041=x^2+10x+25+2016\)
\(=\left(x^2+10x+25\right)+2016\)
\(=\left(x+5\right)^2+2016\ge2016\)
Dấu "=" khi \(x=-5\)
Vậy \(Min=2016\) khi \(x=-5\)
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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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\(Q=3xy\left(x+3y\right)-2xy\left(x+4y\right)-x^2\left(y-1\right)+y^2\left(1-x\right)+36\)\(\Leftrightarrow Q=3x^2y+9xy^2-2x^2y-8xy^2-x^2y+x^2+y^2-xy^2+36\)\(\Leftrightarrow Q=\left(3x^2y-2x^2y-x^2y\right)+\left(9xy^2-8xy^2-xy^2\right)+x^2+y^2+36\)\(\Leftrightarrow Q=x^2+y^2+36\ge36\forall x;y\)
Dấu " = " xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy Min Q là : \(36\Leftrightarrow x=y=0\)
\(D=x+2\sqrt{x-2}+16\)
ĐK: \(x\ge2\)
\(D=x-2+2\sqrt{x-2}+1+17\)
\(D=\left(\sqrt{x-2}+1\right)^2+17\ge1+17=18\)
Vậy Min D = 18 <=> x=2