\(\dfrac{x^2+x+3}{x^2-4}\ge1\Leftrightarrow\dfrac{x^2+x+3}{x^2-4}-1\ge0\)

\(\Leftrightarrow\dfrac{x+7}{x^2-4}\ge0\Rightarrow\left[{}\begin{matrix}-7\le x< -2\\x>2\end{matrix}\right.\)

\(\Rightarrow S\cap\left(-2;2\right)=\varnothing\)

a, \(\left|3x+1\right|>2\)

\(\Leftrightarrow\left(\left|3x+1\right|\right)^2>4\)

\(\Leftrightarrow9x^2+6x+1>4\)

\(\Leftrightarrow9x^2+6x-3>0\)

\(\Leftrightarrow3\left(3x-1\right)\left(x+1\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{3}\\x< -1\end{matrix}\right.\)

b, \(\left|2x-1\right|\le1\)

\(\Leftrightarrow\left(\left|2x-1\right|\right)^2\le1\)

\(\Leftrightarrow4x^2-4x+1\le1\)

\(\Leftrightarrow4x\left(x-1\right)\le0\)

\(\Leftrightarrow0\le x\le1\)

c, ĐK: \(x\ne13\)

\(\left|\dfrac{2}{x-13}\right|>\dfrac{8}{9}\)

\(\Leftrightarrow\dfrac{1}{\left|x-13\right|}>\dfrac{4}{9}\)

\(\Leftrightarrow4\left|x-13\right|< 9\)

\(\Leftrightarrow16\left(x^2-26x+169\right)< 81\)

\(\Leftrightarrow16x^2-416x+2623< 0\)

\(\Leftrightarrow\dfrac{43}{4}< x< \dfrac{61}{4}\)

\(\Rightarrow\) Có hai giả trị thỏa mãn yêu cầu bài toán

a, \(\left|x+2\right|+\left|-2x+1\right|\le x+1\left(1\right)\)

TH1: \(x\le-2\)

\(\Rightarrow x+1\le-1< \left|x+2\right|+\left|-2x+1\right|\)

\(\Rightarrow\) vô nghiệm

TH2: \(-2< x\le\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow x+2-2x+1\le x+1\)

\(\Leftrightarrow x\ge1\)

\(\Rightarrow x\in\left[1;\dfrac{1}{2}\right]\)

TH3: \(x>\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow x+2+2x-1\le x+1\)

\(\Leftrightarrow x\le0\)

\(\Rightarrow\) vô nghiệm

Vậy \(x\in\left[1;\dfrac{1}{2}\right]\)

b, \(\left|x+2\right|-\left|x-1\right|< x-\dfrac{3}{2}\left(2\right)\)

TH1: \(x\le-2\)

\(\left(2\right)\Leftrightarrow-x-2+x-1< x-\dfrac{3}{2}\)

\(\Leftrightarrow x>-\dfrac{3}{2}\)

\(\Rightarrow\) vô nghiệm

TH2: \(-2< x\le1\)

\(\left(2\right)\Leftrightarrow x+2+x-1< x-\dfrac{3}{2}\)

\(\Leftrightarrow x< -\dfrac{5}{2}\)

\(\Rightarrow\) vô nghiệm

TH3: \(x>1\)

\(\left(2\right)\Leftrightarrow x+2-x+1< x-\dfrac{3}{2}\)

\(\Leftrightarrow x>\dfrac{9}{2}\)

\(\Rightarrow x\in\left(\dfrac{9}{2};+\infty\right)\)

Vậy \(x\in\left(\dfrac{9}{2};+\infty\right)\)

c, Tương tự a,b

d, ĐK: \(x\ne-2;x\ne1\)

\(\left|\dfrac{-5}{x+2}\right|< \left|\dfrac{10}{x-1}\right|\)

\(\Leftrightarrow\dfrac{1}{\left|x+2\right|}< \dfrac{2}{\left|x-1\right|}\)

\(\Leftrightarrow2\left|x+2\right|>\left|x-1\right|\)

\(\Leftrightarrow4\left(x+2\right)^2>\left(x-1\right)^2\)

\(\Leftrightarrow4\left(x^2+4x+4\right)>x^2-2x+1\)

\(\Leftrightarrow3x^2+18x+15>0\)

\(\Leftrightarrow...\)

e, ĐK: \(x\ne-1\)

\(\left|\dfrac{2-3\left|x\right|}{1+x}\right|\le1\)

\(\Leftrightarrow\left|2-3\left|x\right|\right|\le\left|x+1\right|\)

\(\Leftrightarrow\left(2-3\left|x\right|\right)^2\le\left(x+1\right)^2\)

\(\Leftrightarrow4+9x^2-12\left|x\right|\le x^2+2x+1\)

\(\Leftrightarrow8x^2-12\left|x\right|-2x+3\le0\)

Đến đây dễ rồi, xét hai trường hợp để phá dấu giá trị tuyệt đối rồi đối chiếu điêì kiện.

a) \(2x-\dfrac{x-3}{5}-4x+1\le0\)

\(\Leftrightarrow10x-x+3-20x+5\le0\)

\(\Leftrightarrow-11x+8\le0\)

\(\Leftrightarrow x\ge\dfrac{8}{11}\)

\(\Rightarrow x\in\left(\dfrac{8}{11};+\infty\right)\)

b) \(\sqrt{x^2+2}\le x-1\)

\(\Leftrightarrow x^2+2\le x^2-2x+1\) \(\left(x-1\ge\sqrt{x^2+2}\ge\sqrt{2}\Rightarrow x\ge1+\sqrt{2}\right)\)

\(\Leftrightarrow x\le-\dfrac{1}{2}\)

\(\Rightarrow x\in\varnothing\)

c) \(\sqrt{x-1}+\sqrt{5-x}+\dfrac{1}{x-3}>\dfrac{1}{x-3}\) (\(x\in\left[1;5\right]\backslash\left\{3\right\}\))

\(\Leftrightarrow\sqrt{x-1}+\sqrt{5-x}>0\)

\(\Leftrightarrow4+2\sqrt{\left(x-1\right)\left(5-x\right)}>0\) ( luôn đúng )

vậy \(x\in\left[1;5\right]\backslash\left\{3\right\}\)

Chọn B

B nhá bạn 

B

a, ĐK: \(x=2017\)

\(\sqrt{x-2017}>\sqrt{2017-x}\)

\(\Leftrightarrow\left\{{}\begin{matrix}2017-x\ge0\\x-2017>2017-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le2017\\x>2017\end{matrix}\right.\)

\(\Rightarrow S=\varnothing\)

b, \(\dfrac{2x^2-3x+4}{x^2+3}>2\)

\(\Leftrightarrow2x^2-3x+4>2x^2+6\)

\(\Leftrightarrow x< -\dfrac{2}{3}\)

\(\Rightarrow S=\left(-\infty;-\dfrac{2}{3}\right)\)

c, ĐK: \(x\le2\)

\(3-2x+\sqrt{2-x}< x+\sqrt{2-x}\)

\(\Leftrightarrow3-2x+\sqrt{2-x}< x+\sqrt{2-x}\)

\(\Leftrightarrow3x>3\)

\(\Leftrightarrow x>1\)

\(\Rightarrow S=(1;2]\)

ĐKXĐ: \(x>\dfrac{1}{2}\)

\(log_{\dfrac{1}{2}}\left(\dfrac{x+1}{2x-1}\right)< 2\)

\(\Rightarrow\dfrac{x+1}{2x-1}>\dfrac{1}{4}\)

\(\Rightarrow x>-\dfrac{5}{2}\)

Kết hợp ĐKXĐ: \(\Rightarrow x>\dfrac{1}{2}\)