Question

[Nguyen phuong nam]

Tim gtnn cua bieu thuc

C=5x^2-7x+4

D=x^2+y^2-2x-4y-6

Answer 1

[Nhóc nhí nhảnh]

\(C=5x^2-7x+4\\ =5\left(x^2-\frac{7}{5}x\right)+4\\ =5\left(x^2-2\cdot x\cdot\frac{7}{10}+\left(\frac{7}{10}\right)^2\right)+\frac{31}{20}\\ =\left(x-\frac{7}{10}\right)^2+\frac{31}{10}\ge\frac{31}{10}\forall x\)

Vậy Min C = \(\frac{31}{10}\)khi \(x=\frac{7}{10}\)

\(D=x^2+y^2-2x-4y-6\\ =\left(x^2-2x+1\right)+\left(y^2-4y+4\right)-11\\ =\left(x-1\right)^2+\left(y-2\right)^2-11\)

Ta thấy \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(y-2\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow D=\left(x-1\right)^2+\left(y-2\right)^2-11\ge-11\forall x,y\)

Vậy min D = -11 khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Answer 2

[Lân Trần Quốc]

\(C=5x^2-7x+4\\ =5x^2-7x+\frac{49}{20}+\frac{31}{20}\\ =\left(x\sqrt{5}-\frac{7\sqrt{5}}{10}\right)^2+\frac{31}{20}\ge\frac{31}{20}\left(\forall x\in R\right)\)

Đẳng thức xảy ra \(\Leftrightarrow x\sqrt{5}-\frac{7\sqrt{5}}{10}=0\Leftrightarrow\sqrt{5}\left(x-\frac{7}{10}\right)=0\Leftrightarrow x=\frac{7}{10}\)

\(D=x^2+y^2-2x-4y-6=0\\ =x^2-2x+1+y^2-4y+4-11\\ =\left(x-1\right)^2+\left(y-2\right)^2-11\ge-11\left(\forall x,y\in R\right)\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Vậy \(minC=\frac{31}{20}\), đạt được khi \(x=\frac{7}{10}\); và \(minD=-11\), đạt được khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Chúc bạn học tốt nha.