Question

Tìm GTNN của biểu thức:

A=x-\(\sqrt{x-2020}\)

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

C=\(\sqrt{x^2+10x+25}+\sqrt{x^2-6x+9}\)

D=x(x+1)(x+2)(x+3)

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

F=\(\frac{x^2}{x^4+4}\)

Answer 1

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

= | x+1 | - | 1-x | \(\ge\left|x+1+1-x\right|=\left|2\right|=2\)

dấu "=" xảy ra <=> \(\left(x+1\right)\left(1-x\right)\ge0\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\1-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\1-x\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-1\\x\le1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\end{matrix}\right.\)

<=> \(-1\le x\le1\)

Vậy min C = 1 khi và chỉ khi \(-1\le x\le1\)