Question

Tìm GTNN

C=\(^{2X^2-5X+1}\)

D=\(^{x^2+2x+y^2-8y-4}\)

Tìm GTLN

C=\(-2x^2+2x-1\)

D=\(-x^2-y^2-x+y-4\)

Answer 1

Tìm GTNN 

Câu 1 :

\(C=2x^2-5x+1\)

\(C=2\left(x^2-\frac{5}{2}x+\frac{1}{2}\right)\)

\(C=2\left(x^2-2\cdot x\cdot\frac{5}{4}+\frac{25}{16}-\frac{17}{16}\right)\)

\(C=2\left[\left(x-\frac{5}{4}\right)^2-\frac{17}{16}\right]\)

\(C=2\left(x-\frac{5}{4}\right)^2-\frac{17}{8}\ge\frac{-17}{8}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-\frac{5}{4}=0\Leftrightarrow x=\frac{5}{4}\)

Câu 2 : 

\(D=x^2+2x+y^2-8y-4\)

\(D=x^2+2\cdot x\cdot1+1^2+y^2-2\cdot y\cdot4+4^2-21\)

\(D=\left(x+1\right)^2+\left(y-2\right)^2-21\ge-21\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+1=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}}\)

Tìm GTLN :

Câu 1 :

\(C=-2x^2+2x-1\)

\(C=-2\left(x^2-x+\frac{1}{2}\right)\)

\(C=-2\left(x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\right)\)

\(C=-2\left[\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\right]\)

\(C=-2\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\)

\(C=-\frac{1}{2}-2\left(x-\frac{1}{2}\right)^2\le-\frac{1}{2}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)

Answer 2

Câu 2 :

\(D=-x^2-y^2-x+y-4\)

\(D=-\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)-\left(y^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)-\frac{7}{2}\)

\(D=-\left(x+\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2-\frac{7}{2}\)

\(D=\frac{-7}{2}-\left[\left(x+\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2\right]\le\frac{-7}{2}\forall x;y\)

Dấu "=' xảy ra \(\Leftrightarrow\hept{\begin{cases}x+\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{1}{2}\end{cases}}}\)

Answer 3

Câu 1) \(C=2x^2-5x+1\)

\(=2x^2-5x+\frac{25}{8}-\frac{17}{8}\)

\(=2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)-\frac{17}{8}\)

\(=2\left(x-\frac{5}{4}\right)^2-\frac{17}{8}\ge-\frac{17}{8}\)(do \(2\left(x-\frac{5}{4}\right)^2\ge0\forall x\))

Dấu "=" xảy ra \(\Leftrightarrow2\left(x-\frac{5}{4}\right)^2=0\Leftrightarrow x=\frac{5}{4}\)

Vậy \(C_{min}=-\frac{17}{8}\Leftrightarrow x=\frac{5}{4}\)

Câu 2) \(C=-2x^2+2x-1\)

\(=-2x^2+2x-\frac{1}{2}-\frac{1}{2}\)

\(=-2\left(x^2-1x+\frac{1}{4}\right)-\frac{1}{2}\)

\(=-2\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\le-\frac{1}{2}\) ( do \(-2\left(x-\frac{1}{2}\right)^2\le0\forall x\))

Dấu "=" xãy ra \(\Leftrightarrow-2\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x=\frac{1}{2}\)

Vậy \(C_{max}=-\frac{1}{2}\Leftrightarrow x=\frac{1}{2}\)