ĐKXĐ: \(\left\{{}\begin{matrix}4x-20\ne0\\50-2x^2\ne0\\6x+30\ne0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}4x-20\ne0\\x^2-25\ne0\\6x+30\ne0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x-5\ne0\\x+5\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)

=> \(x\ne\pm5\)

Ta có : \(\frac{3}{4x-20}+\frac{15}{50-2x^2}+\frac{7}{6x+30}=0\)

=> \(\frac{3}{4\left(x-5\right)}-\frac{15}{2\left(x-5\right)\left(x+5\right)}+\frac{7}{6\left(x+5\right)}=0\)

=> \(\frac{9\left(x+5\right)}{12\left(x^2-25\right)}-\frac{90}{12\left(x^2-25\right)}+\frac{14\left(x-5\right)}{12\left(x^2-25\right)}=0\)

=> \(9\left(x+5\right)-90+14\left(x-5\right)=0\)

=> \(9x+45-90+14x-70=0\)

=> \(23x=115\)

=> \(x=5\) ( KTM )

Vậy phương trình vô nghiệm .

\(\Leftrightarrow\dfrac{3}{4\left(x-5\right)}-\dfrac{15}{2\left(x-5\right)\left(x+5\right)}+\dfrac{7}{6\left(x+5\right)}=0\)

\(\Leftrightarrow3\cdot3\left(x+5\right)-15\cdot6+7\cdot2\cdot\left(x-5\right)=0\)

=>9x+45-90+14x-70=0

=>23x-115=0

hay x=5(loại)

a) Để biểu thức vô nghĩa thì \(\dfrac{3x-2}{5}-\dfrac{x-4}{3}=0\)

\(\Leftrightarrow\dfrac{3x-2}{5}=\dfrac{x-4}{3}\)

\(\Leftrightarrow3\left(3x-2\right)=5\left(x-4\right)\)

\(\Leftrightarrow9x-6=5x-20\)

\(\Leftrightarrow9x-5x=-20+6\)

\(\Leftrightarrow4x=-14\)

\(\Leftrightarrow x=-\dfrac{7}{2}\)

\(\text{a) }\left|2-5x\right|=\left|3x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}2-5x=3x+1\\2-5x=-3x-1\end{matrix}\right. \Leftrightarrow\left[{}\begin{matrix}-5x-3x=1-2\\-5x+3x=-1-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-8x=-1\\-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{8}\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy tập nghiệm phương trình là \(S=\left\{\dfrac{1}{8};\dfrac{3}{2}\right\}\)

\(\text{b) }\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\)

ĐXKĐ của phương trình \(:x\ne\pm5\)

\(\text{Ta có }:\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}+\dfrac{15}{2\left(25-x^2\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}-\dfrac{15}{2\left(x+5\right)\left(x-5\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{9\left(x+5\right)}{12\left(x+5\right)\left(x-5\right)}-\dfrac{90}{12\left(x+5\right)\left(x-5\right)}+\dfrac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}=0\\ \Rightarrow9x+45-90+14x-70=0\\ \Leftrightarrow23x=115\\ \Leftrightarrow x=5\left(KTM\right)\)

Vậy phương trình vô nghiệm

\(\text{c) }\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\\ \Leftrightarrow\left(\dfrac{x+29}{31}+1\right)-\left(\dfrac{x+27}{33}+1\right)=\left(\dfrac{x+17}{43}+1\right)-\left(\dfrac{x+15}{45}+1\right)\\ \Leftrightarrow\dfrac{x+60}{31}-\dfrac{x+60}{33}-\dfrac{x+60}{43}+\dfrac{x+60}{45}=0\\ \Leftrightarrow\left(x+60\right)\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\right)=0\\ \Leftrightarrow x+60=0\left(\text{Vì }\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\ne0\right)\\ \Leftrightarrow x=-60\)

Vậy \(x=-60\) là nghiệm của phương trình

b: Đặt \(x^2-6x-2=a\)

Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)

=>(a+2)(a+7)=0

\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)

=>x(x-6)(x-1)(x-5)=0

hay \(x\in\left\{0;1;6;5\right\}\)

c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)

\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)

\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)

\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)

=>26x=-3

hay x=-3/26

24: 

\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)

\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)

=>(x+10)(x-2)=0

=>x=-10 hoặc x=2

25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)

\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)

\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)

\(\Leftrightarrow x+5=0\)

hay x=-5

\(\Leftrightarrow\dfrac{-7}{x^2+3x-10}+\dfrac{x+4}{x+5}+\dfrac{x+3}{x-2}+3=0\)

\(\Leftrightarrow-7+x^2+2x-8+x^2+8x+15+3x^2+9x-30=0\)

\(\Leftrightarrow5x^2+19x-30=0\)

hay \(x\in\left\{\dfrac{6}{5}\right\}\)

1) -2/3

1: \(\Leftrightarrow3x+4=2\)

=>3x=-2

=>x=-2/3

2: \(\Leftrightarrow7x-7=6x-30\)

=>x=-23

3: =>\(5x-5=3x+9\)

=>2x=14

=>x=7

4: =>9x+15=14x+7

=>-5x=-8

=>x=8/5