a) Ta có: (5x-1)(x-3)<0

nên 5x-1 và x-3 trái dấu

Trường hợp 1:

\(\left\{{}\begin{matrix}5x-1>0\\x-3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x< 3\end{matrix}\right.\Leftrightarrow\dfrac{1}{5}< x< 3\)

Trường hợp 2:

\(\left\{{}\begin{matrix}5x-1< 0\\x-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x>3\end{matrix}\right.\Leftrightarrow loại\)

Vậy: S={x|\(\dfrac{1}{5}< x< 3\)}

b: =>1/4x+4/5-x-5=1/3x+1-1/2x+1

=>-3/4x+1/6x=2+5-4/5=24/5

=>x=-288/35

c: =>6x^2+3x-30x-15=6x^2+10x-21x-35

=>-27x-15=-11x-35

=>-16x=-20

=>x=5/4

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

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

a) \(2\chi-3=3\left(\chi+1\right)\)

\(\Leftrightarrow2\chi-3=3\chi+3\)

\(\Leftrightarrow2\chi-3\chi=3+3\)

\(\Leftrightarrow\chi=-6\)

Vậy phương trình có tập nghiệm S= \(\left\{-6\right\}\)

\(3\chi-3=2\left(\chi+1\right)\)

\(\Leftrightarrow3\chi-3=2\chi+2\)

\(\Leftrightarrow3\chi-2\chi=2+3\)

\(\Leftrightarrow\chi=5\)

Vậy phương trình có tập nghiệm S= \(\left\{5\right\}\)

b) \(\left(3\chi+2\right)\left(4\chi-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3\chi+2=0\\4\chi-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3\chi=-2\\4\chi=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\chi=\dfrac{-2}{3}\\\chi=\dfrac{5}{4}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S= \(\left\{\dfrac{-2}{3};\dfrac{5}{4}\right\}\)

\(\left(3\chi+5\right)\left(4\chi-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3\chi+5=0\\4\chi-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3\chi=-5\\4\chi=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\chi=\dfrac{-5}{3}\\\chi=\dfrac{1}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S= \(\left\{\dfrac{-5}{3};\dfrac{1}{2}\right\}\)

c) \(\left|\chi-7\right|=2\chi+3\)

Trường hợp 1: 

Nếu \(\chi-7\ge0\Leftrightarrow\chi\ge7\)

Khi đó:\(\left|\chi-7\right|=2\chi+3\)

 \(\Leftrightarrow\chi-7=2\chi+3\)

\(\Leftrightarrow\chi-2\chi=3+7\)

\(\Leftrightarrow\chi=-10\) (KTMĐK)

Trường hợp 2:

Nếu \(\chi-7\le0\Leftrightarrow\chi\le7\)

Khi đó: \(\left|\chi-7\right|=2\chi+3\)

\(\Leftrightarrow-\chi+7=2\chi+3\)

\(\Leftrightarrow-\chi-2\chi=3-7\)

\(\Leftrightarrow-3\chi=-4\)

\(\Leftrightarrow\chi=\dfrac{4}{3}\)(TMĐK)

Vậy phương trình có tập nghiệm S=\(\left\{\dfrac{4}{3}\right\}\)

\(\left|\chi-4\right|=5-3\chi\)

Trường hợp 1:  

Nếu \(\chi-4\ge0\Leftrightarrow\chi\ge4\)

Khi đó: \(\left|\chi-4\right|=5-3\chi\)

\(\Leftrightarrow\chi-4=5-3\chi\)

\(\Leftrightarrow\chi+3\chi=5+4\)

\(\Leftrightarrow4\chi=9\)

\(\Leftrightarrow\chi=\dfrac{9}{4}\)(KTMĐK)

Trường hợp 2: Nếu \(\chi-4\le0\Leftrightarrow\chi\le4\)

Khi đó: \(\left|\chi-4\right|=5-3\chi\)

\(\Leftrightarrow-\chi+4=5-3\chi\)

\(\Leftrightarrow-\chi+3\chi=5-4\)

\(\Leftrightarrow2\chi=1\)

\(\Leftrightarrow\chi=\dfrac{1}{2}\)(TMĐK)

Vậy phương trình có tập nghiệm S=\(\left\{\dfrac{1}{2}\right\}\)

a) \(\left(x-2\right)^2=\left(x-4\right)\left(x+4\right)\) 

\(\Leftrightarrow x^2-4x+4-x^2+16=0\)

\(\Leftrightarrow20-4x=0\)

\(\Leftrightarrow4x=20\)

\(\Leftrightarrow x=5\)

Vậy S = {5}

b) ĐKXĐ: \(x\ne0;x\ne-2\)

\(\dfrac{x+2}{x}=\dfrac{\left(x+1\right)\left(x+4\right)}{x^2+2x}+\dfrac{x}{x+2}\)

\(\Leftrightarrow\dfrac{x+2}{x}=\dfrac{x^2+4x+x+4+x^2}{x\left(x+2\right)}\)

\(\Leftrightarrow\dfrac{x+2}{x}=\dfrac{2x^2+5x+4}{x\left(x+2\right)}\)

\(\Rightarrow x\left(x+2\right)^2=x\left(2x^2+5x+4\right)\)

\(\Leftrightarrow x^3+4x^2+4x=2x^3+5x^2+4x\)

\(\Leftrightarrow x^3+x^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(TM\right)\end{matrix}\right.\)

Vậy S = {-1}

c) Câu này mình không chắc về đề lắm! Bạn dùng ô chữ M bị ngược để viết lại đề nhé!

a) Ta có: \(\left(x-2\right)^2=\left(x-4\right)\left(x+4\right)\)

\(\Leftrightarrow x^2-4x+4=x^2-16\)

\(\Leftrightarrow x^2-4x+4-x^2+16=0\)

\(\Leftrightarrow-4x+20=0\)

\(\Leftrightarrow-4x=-20\)

hay x=5

Vậy: S={5}

\(\text{2x - (x - 3)(5 - x) = (x+4)}^2.\)

\(\Leftrightarrow2x-\left(5x-x^2-15+3x\right)=x^2+8x+16.\)

\(\Leftrightarrow2x-5x+x^2+15-3x-x^2-8x-16=0.\)

\(\Leftrightarrow-14x-1=0.\Leftrightarrow x=\dfrac{-1}{14}.\)

\(\text{(4x + 1)(x - 2) + 25 = (2x+3)}^2-4x.\)

\(\Leftrightarrow4x^2-8x+x-2+25=4x^2+12x+9-4x.\)

\(\Leftrightarrow-15x+14=0.\Leftrightarrow x=\dfrac{14}{15}.\)

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: \(S=\left\{-\dfrac{7}{9}\right\}\)