Question

Tìm GTNN của các biểu thức sau:

F=\(\left|2x-2\right|+\left|2x-2003\right|\)

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

H=\(\left|x-2018\right|+\left|x-2019\right|+\left|x-2020\right|\)

CÁC BẠN GIÚP MÌNH VỚI !!!!!!!!!!!!!!!!!

Answer 1

Lời giải:

Bạn áp dụng BĐT sau:

$|a|+|b|\geq |a+b|$. Dấu "=" xảy ra khi $ab\geq 0$

Ta có:

\(F=|2x-2|+|2x-2003|=|2x-2|+|2003-2x|\geq |2x-2+2003-2x|=2001\)

Vậy $F_{\min}=2001$. Dấu "=" xảy ra khi $(2x-2)(2003-2x)\geq 0$

$\Leftrightarrow 1\leq x\leq \frac{2003}{2}$

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\(G=|2x-3|+\frac{1}{2}|4x-1|=|2x-3|+|2x-\frac{1}{2}|=|3-2x|+|2x-\frac{1}{2}|\geq |3-2x+2x-\frac{1}{2}|\)

\(=\frac{5}{2}\)

Vậy $G_{\min}=\frac{5}{2}$. Dấu "=" xảy ra khi $(3-2x)(2x-\frac{1}{2})\geq 0$

$\Leftrightarrow \frac{1}{4}\leq x\leq \frac{3}{2}$

Answer 2

$H=|x-2018|+|x-2019|+|x-2020|$

$=|x-2018|+|x-2020|+|x-2019|=|x-2018|+|2020-x|+|x-2019|$

Ta có:

$|x-2018|+|2020-x|\geq |x-2018+2020-x|=2$

$|x-2019|\geq 0$ với mọi $x$

$\Rightarrow H\geq 2$

Vậy $H_{\min}=2$. Dấu "=" xảy ra khi \(\left\{\begin{matrix} (x-2018)(2020-x)\geq 0\\ x-2019=0\end{matrix}\right.\Leftrightarrow x=2019\)