Tìm GTNN của biểu thức D : (2x^2 + x +1)/ -2
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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(D=\left(2x-1\right)^2+\left(x+2\right)^2+12\)
\(=4x^2-4x+1+x^2+4x+4+12\)
\(=5x^2+17\ge17\forall x\)
Dấu '=' xảy ra khi x=0
\(D=\left(2x-1\right)^2+\left(x+2\right)^2+12\)
\(=4x^2-4x+1+x^2+4x+4+12=5x^2+17\ge17\)
Dấu ''='' xảy ra khi x = 0
Vậy GTNN D bằng 17 tại x = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
c: Ta có: \(\left(x+1\right)^2\ge0\forall x\)
\(\left(y-\dfrac{1}{3}\right)^2\ge0\forall y\)
Do đó: \(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2\ge0\forall x,y\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\forall x,y\)
Dấu '=' xảy ra khi x=-1 và \(y=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^2-6x+10\)
\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)
\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\) \(\forall x\in z\)
\(\Leftrightarrow A_{min}=1khix=3\)
\(B=3x^2-12x+1\)
\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)
\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\) \(\forall x\in z\)
\(\Leftrightarrow B_{min}=-11khix=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm GTLN:
\(A=-x^2+6x-15\)
\(=-\left(x^2-6x+15\right)\)
\(=-\left(x^2-2.x.3+9+6\right)\)
\(=-\left(x+3\right)^2-6\le0\forall x\)
Dấu = xảy ra khi:
\(x-3=0\Leftrightarrow x=3\)
Vậy Amax = - 6 tại x = 3
Tìm GTNN :
\(A=x^2-4x+7\)
\(=x^2+2.x.2+4+3\)
\(=\left(x+2\right)^2+3\ge0\forall x\)
Dấu = xảy ra khi:
\(x+2=0\Leftrightarrow x=-2\)
Vậy Amin = 3 tại x = - 2
Các câu còn lại làm tương tự nhé... :)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(P=\dfrac{2x^2+x}{\left(x+1\right)^2}\Rightarrow P+\dfrac{1}{4}=\dfrac{9x^2+6x+1}{4\left(x+1\right)^2}=\dfrac{\left(3x+1\right)^2}{4\left(x+1\right)^2}\ge0\).
Dấu "=" xảy ra khi và chỉ khi \(x=-\dfrac{1}{3}\).
Vậy..
Ta có: \(2x^2+x+1\)
\(=\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.\frac{1}{2\sqrt{2}}+\frac{1}{8}+\frac{7}{8}\)
\(=\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2+\frac{7}{8}\ge\frac{7}{8}\)
\(\frac{\Rightarrow\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2+\frac{7}{8}}{-2}\le\frac{-7}{16}\)
(Dấu "="\(\Leftrightarrow\sqrt{2}x+\frac{1}{2\sqrt{2}}=0\Leftrightarrow x=\frac{-1}{4}\)
\(D=\frac{2x^2+x+1}{-2}\)
\(=\frac{2\left(x^2+\frac{1}{2}x+\frac{1}{2}\right)}{-2}\)
\(=\frac{2\left(x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}+\frac{1}{2}\right)}{-2}\)
\(=\frac{2\left(x+\frac{1}{2}\right)^2+\frac{7}{8}}{-2}\)
Vì \(2\left(x+\frac{1}{2}\right)^2\ge0;\forall x\)
\(\Rightarrow2\left(x+\frac{1}{2}\right)^2+\frac{7}{8}\ge\frac{7}{8};\forall x\)
\(\Rightarrow\frac{2\left(x+\frac{1}{2}\right)^2+\frac{7}{8}}{-2}\ge\frac{-7}{16};\forall x\)
Dấu'="xảy ra \(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy \(D_{min}=\frac{-7}{16}\)\(\Leftrightarrow x=\frac{-1}{2}\)