a) x={-2017;-2016;-2015;-2014;-2013;......;2015;2016;2017;2018}

xin lỗi bạn Thái Yrần Thảo Vy nha do mình quên ghi đề

tổng =2018 nha

a)Ta có: \(\dfrac{x+3}{x+1}+\dfrac{1}{3}\ge0\)

\(\Leftrightarrow\dfrac{3x+9+x+1}{3\left(x+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{4x+10}{3x+3}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1>0\\4x+10\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-1\\x\le-\dfrac{5}{2}\end{matrix}\right.\)

b) Ta có: \(\dfrac{x+2}{x+3}+\dfrac{1}{3}\le0\)

\(\Leftrightarrow\dfrac{3x+6+x+3}{3\left(x+3\right)}\le0\)

\(\Leftrightarrow\dfrac{4x+9}{3x+9}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+9>0\\4x+9\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\le-\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow-3< x\le-\dfrac{9}{4}\)

a)\(\dfrac{x+3}{x+1}\ge-\dfrac{1}{3}\left(x\ne-1\right)\)

\(\Leftrightarrow\dfrac{x+3}{x+1}+\dfrac{1}{3}\ge0\)

\(\Leftrightarrow\dfrac{3x+9+x+1}{3x+3}\ge0\)

\(\Leftrightarrow\dfrac{4x+10}{3x+3}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x+10\ge0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4x+10\le0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{5}{2}\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{-5}{2}\\x< -1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>-1\\x\le\dfrac{-5}{2}\end{matrix}\right.\)

 b) \(\dfrac{x+2}{x+3}\le-\dfrac{1}{3}\left(x\ne-3\right)\)

\(\dfrac{x+2}{x+3}+\dfrac{1}{3}\le0\)

\(\Leftrightarrow\dfrac{3x+6+x+3}{3x+9}\le0\)

\(\Leftrightarrow\dfrac{4x+9}{3x+9}\le0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x+9\ge0\\3x+9< 0\end{matrix}\right.\\\left\{{}\begin{matrix}4x+9\le0\\3x+9>0\end{matrix}\right.\end{matrix}\right.\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{9}{4}\\x< -3\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-\dfrac{9}{4}\\x>-3\end{matrix}\right.\end{matrix}\right.\)    

TH1: loại

TH2: TM

Vậy no của BPT là :\(-\dfrac{9}{4}\ge x>-3\)

chúc bạn học tốt

a) \(2-x\ge0\Leftrightarrow x\le2\)

b) \(2+x\ge0\Leftrightarrow x\ge-2\)

c) \(7-x\ge0\Leftrightarrow x\le7\)

chờ x,y,z dương thỏa mãn ( x + y lớn hơn hoặc bằng  3)CMR : x + y + 1/(2x) + 2/y lớn hơn hoặc bằng 9Giúp mình với, mình đang rất gấp, sẽ hậu tạ sau nha, bạn muốn gì cũng được. OK!!!!

Ý quên là " lớn hơn hoặc bằng 9/2"

Ta có: \(\left|x+20\right|;\left|y-11\right|;\left|z+2003\right|\ge0\)

\(\Rightarrow\left|x+20\right|+\left|y-11\right|+\left|z+2003\right|\ge0\)

Theo đề: \(\left|x+20\right|+\left|y-11\right|+\left|z+2003\right|\le0\)

\(\Rightarrow\left|x+20\right|+\left|y-11\right|+\left|z+2003\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|x+20\right|=0\\\left|y-11\right|=0\\\left|z+2003\right|=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-20\\y=11\\z=-2003\end{cases}}\)

có phải là thế nay ko bạn\(a=-3\le a\le0\)

Khi  \(-3\le a\le0\Leftrightarrow\left|a+3\right|=a+3\)

                                        \(\left|a-3\right|=3-a\)

\(pt\Leftrightarrow a+3+3-a\)

\(=\left(a-a\right)+\left(3+3\right)\)

\(=0+6\)

\(=6\)

Vậy ...