Question

Bài 1: Tìm già trị nhỏ nhất :

A=\(\left|2x-\frac{1}{3}\right|-1\frac{3}{4}\)

B=\(\frac{1}{3}\left|x-2\right|+2\left|3-\frac{1}{2}y\right|+4\)

Answer 1

\(A=\left|2x-\frac{1}{3}\right|-1\frac{3}{4}\)

Có: \(\left|2x-\frac{1}{3}\right|\ge0\text{ với mọi }x\)

\(\Rightarrow\left|2x-\frac{1}{3}\right|-1\frac{3}{4}\ge-1\frac{3}{4}\text{ với mọi }x\)

\(\Rightarrow A\ge-1\frac{3}{4}\text{ với mọi }x\)

Vậy GTNN của \(A=-1\frac{3}{4}\)

Dấu '"=" xảy ra khi \(\left|2x-\frac{1}{3}\right|=0\\ \Leftrightarrow2x-\frac{1}{3}=0\\ \Leftrightarrow2x=\frac{1}{3}\\ \Leftrightarrow x=\frac{1}{6}\)

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\(B=\frac{1}{3}\left|x-2\right|+2\left|3-\frac{1}{2}y\right|+4\)

Có: \(\left\{{}\begin{matrix}\frac{1}{3}\left|x-2\right|\ge0\text{ với mọi }x\\2\left|3-\frac{1}{2}y\right|\ge0\text{ với mọi }y\end{matrix}\right.\)

\(\Rightarrow\frac{1}{3}\left|x-2\right|+2\left|3-\frac{1}{2}y\right|\ge0\text{ với mọi }x;y\\ \Rightarrow\frac{1}{3}\left|x-2\right|+2\left|3-\frac{1}{2}y\right|+4\ge4\text{ với mọi }x;y\\ \Rightarrow B\ge4\text{ với mọi }x;y\)

Vậy GTNN của B = 4

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\frac{1}{3}\left|x-2\right|=0\\2\left|3-\frac{1}{2}y\right|=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left|x-2\right|=0\\\left|3-\frac{1}{2}y\right|=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3-\frac{1}{2}y=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\\frac{1}{2}y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)

Answer 2

làm đại:((,sai thì thông cảm nha :))

Answer 3

khúc cuối sai sai tý kkk

Answer 4

\(A=\left|2x-\frac{1}{3}\right|-1\frac{3}{4}=\left|2x-\frac{1}{3}\right|+\left(-\frac{7}{4}\right)\)

+Có: \(\left|2x-\frac{1}{3}\right|\ge0với\forall x\\ \Rightarrow\left|2x-\frac{1}{3}\right|+\left(-\frac{7}{4}\right)\ge-\frac{7}{4}\\ \Leftrightarrow A\ge-\frac{7}{4}\)

+Dấu ''='' xảy ra khi \(\left|2x-\frac{1}{3}\right|=0\Leftrightarrow x=\frac{1}{6}\)

+Vậy \(A_{min}=-\frac{7}{4}\) khi \(x=\frac{1}{6}\)

Answer 5

\(B=\frac{1}{3}\left|x-2\right|+2\left|3-\frac{1}{2}y\right|+4\)

+Có: \(\left|x-2\right|\ge0với\forall x\\ \left|3-\frac{1}{2}y\right|\ge0với\forall x\\ \Rightarrow\frac{1}{3}\left|x-2\right|+2\left|3-\frac{1}{2}y\right|+4\ge4\\ \Leftrightarrow B\ge4\)

+Dấu ''='' xảy ra khi \(\left|x-2\right|=0\Leftrightarrow x=2;\left|3-\frac{1}{2}y\right|=0\Leftrightarrow y=6\)

+Vậy \(B_{min}=4\) khi \(x=2;\left|3-\frac{1}{2}y\right|=0\Leftrightarrow y=6\)