Question

Giúp với

Answer 1

Ta có: \(D=\left|x+1,5\right|+\left|x-2\right|=\left|x+1,5\right|+\left|2-x\right|\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(D=\left|x+1,5\right|+\left|2-x\right|\ge\left|x+1,5+2-x\right|=\left|3,5\right|=3,5\)

Dấu " = " xảy ra khi \(\left\{{}\begin{matrix}x+1,5\ge0\\2-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge-1,5\\x\le2\end{matrix}\right.\Rightarrow-1,5\le x\le2\)

Vậy \(MIN_D=3,5\) khi \(-1,5\le x\le2\)

Answer 2

\(\left\{-1;0;1;2\right\}\)

Answer 3

Ta có: \(D=\left|x+1,5\right|+\left|x-2\right|=\left|x+1,5\right|+\left|2-x\right|\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(D=\left|x+1,5\right|+\left|2-x\right|\ge\left|x+1,5+2-x\right|=\left|3,5\right|=3,5\)

Dấu " = " xảy ra khi \(\left\{{}\begin{matrix}x+1,5\ge0\\2-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge-1,5\\x\le2\end{matrix}\right.\Rightarrow-1,5\le x\le2\)

Mà \(x\in Z\)

\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)

Vậy \(MIN_D=3,5\) khi \(x\in\left\{-1;0;1;2\right\}\)

Answer 4

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