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\(D=x+2\sqrt{x-2}+16\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x=-5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=x^2+y^2-x+6y+10\)
\(=\left(x-\frac{1}{2}\right)^2+\left(y-3\right)^2+\frac{3}{4}\)
\(\left(x-\frac{1}{2}\right)^2\ge0;\)\(\left(y-3\right)^2\ge0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-3\right)^2\ge0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-3\right)^2+\frac{3}{4}\ge0+\frac{3}{4}=\frac{3}{4}\)
\(\Rightarrow A\ge\frac{3}{4}\)
Dấu"=" xảy ra khi \(\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x=\frac{1}{2};\left(y-3\right)^2=0\Leftrightarrow y=3\)
Vậy \(MinA=\frac{3}{4}\Leftrightarrow x=\frac{1}{2};y=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Q=3xy\left(x+3y\right)-2xy\left(x+4y\right)-x^2\left(y-1\right)+y^2\left(1-x\right)+36\)\(\Leftrightarrow Q=3x^2y+9xy^2-2x^2y-8xy^2-x^2y+x^2+y^2-xy^2+36\)\(\Leftrightarrow Q=\left(3x^2y-2x^2y-x^2y\right)+\left(9xy^2-8xy^2-xy^2\right)+x^2+y^2+36\)\(\Leftrightarrow Q=x^2+y^2+36\ge36\forall x;y\)
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 : \(36\Leftrightarrow x=y=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ý bạn là \(\sqrt{\left(x+\sqrt{x+81}\right)}-10\) á
ừ.