\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)

Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)

a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)

b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)

\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)

\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)

\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)

\(\Leftrightarrow x\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy: S={0;6}

c) Ta có: \(3x-15=2x\left(x-5\right)\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)

d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)

\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)

\(\Leftrightarrow30-6x=6x-8\)

\(\Leftrightarrow30-6x-6x+8=0\)

\(\Leftrightarrow-12x+38=0\)

\(\Leftrightarrow-12x=-38\)

\(\Leftrightarrow x=\dfrac{19}{6}\)

Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)

e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow6x+4-3x-1=12x+10\)

\(\Leftrightarrow3x+3-12x-10=0\)

\(\Leftrightarrow-9x-7=0\)

\(\Leftrightarrow-9x=7\)

\(\Leftrightarrow x=-\dfrac{7}{9}\)

Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)

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.

1: \(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)

=>-4x<12

hay x>-3

2: \(\Leftrightarrow6+2x+2>2x-1-12\)

=>8>-13(đúng)

4: \(\dfrac{2x+1}{x-3}\le2\)

\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}< =0\)

=>x-3<0

hay x<3

6: =>(x+4)(x-1)<=0

=>-4<=x<=1

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

Chọn B

B nhá bạn 

B

\(\dfrac{2x-1}{x+1}-2< 0.\left(x\ne-1\right).\\ \Leftrightarrow\dfrac{2x-1-2x-2}{x+1}< 0.\Leftrightarrow\dfrac{-3}{x+1}< 0.\)

Mà \(-3< 0.\)

\(\Rightarrow x+1>0.\Leftrightarrow x>-1\left(TMĐK\right).\)

\(\dfrac{x^2-2x+5}{x-2}-x+1\ge0.\left(x\ne2\right).\\ \Leftrightarrow\dfrac{x^2-2x+5-x^2+2x+x-2}{x-2}\ge0.\\ \Leftrightarrow\dfrac{x+3}{x-2}\ge0.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0.\\x-2\ge0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0.\\x-2\le0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3.\\x\ge2.\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3.\\x\le2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge2.\\x\le-3.\end{matrix}\right.\)

Kết hợp ĐKXĐ.

\(\Rightarrow\left[{}\begin{matrix}x>2.\\x\le-3.\end{matrix}\right.\)

\(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}\le0.\left(x\ne1;x\ne\dfrac{-3}{2}\right).\)

Đặt \(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}=f\left(x\right).\)

Ta có bảng sau:

Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left(\dfrac{-3}{2};\dfrac{-1}{2}\right)\cup\)(1;2].

2) −(x2−2x+5)x−2−x+1≥0−(x2−2x+5)x−2−x+1≥0                                                        ⇔$-{}^{}$