a, \(\Rightarrow x-5\ne0\Leftrightarrow x\ne5\)

b, mà \(\left(5x-3\right)^2\ge0\forall x\)

 \(\Rightarrow5x-3=0\Leftrightarrow x=\dfrac{3}{5}\)

(x + 1) + (x + 2)  + (x + 3) = 4x

x + 1 + x + 2 + x + 3 = 4x

(x + x + x) + (1 + 2 + 3) = 4x

3x + 6 = 4x

3x - 4x = -6

- x = -6

   x = \(\frac{-6}{-1}\)

   x = 6

\(\frac{x}{3}-x+1=\frac{x}{4}-\frac{x+1}{3}\)

\(\Leftrightarrow12.\left(\frac{x}{3}-x+1\right)=12.\left(\frac{x}{4}-\frac{x+1}{3}\right)\)

\(\Leftrightarrow4x-12x+12=3x-4.\left(x+1\right)\)

\(\Leftrightarrow4x-12x+12=3x-4x-4\)

\(\Leftrightarrow4x-12x-3x+4x=\left(-12\right)-4\)

\(\Leftrightarrow\left(-7\right)x=-16\)

\(\Leftrightarrow x=\frac{16}{7}\)

Vậy \(x=\frac{16}{7}\)

\(4x-12x+12=3x-4\left(x+1\right)\Leftrightarrow-8x+12=-x-4\)

\(\Leftrightarrow-7x=-16\Leftrightarrow x=\dfrac{16}{7}\)

sửa đề 

\(\left(x+4\right)\left(x^2+\dfrac{1}{2x-1,5}\right)-\left(3-x\right)\left(x^2+\dfrac{1}{2x-1,5}\right)=0\)

đk : x khác 3/4 

\(\Leftrightarrow\left(x+4+x-3\right)\left(x^2+\dfrac{1}{2x-1,5}\right)=0\)

\(\Leftrightarrow\left(x^2+1>0\right)\left(\dfrac{2x^3-1,5x+1}{2x-1,5}\right)=0\Rightarrow2x^3-1,5x+1=0\Leftrightarrow x=-1,0979...\)

a,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne2\\x\ne5\end{matrix}\right.\)

\(\dfrac{3x}{x-2}+\dfrac{3x}{\left(x-2\right)\left(x-5\right)}=\dfrac{x}{x-5}\\ \Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}+\dfrac{3x}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=0\\ \Leftrightarrow x\left(\dfrac{3x-15}{\left(x-2\right)\left(x-5\right)}+\dfrac{3}{\left(x-2\right)\left(x-5\right)}-\dfrac{x-2}{\left(x-2\right)\left(x-5\right)}\right)=0\\ \Leftrightarrow x.\dfrac{3x-15+3-x+2}{\left(x-2\right)\left(x-5\right)}=0\\ \Leftrightarrow\dfrac{x\left(2x-10\right)}{\left(x-2\right)\left(x-5\right)}=0\\ \Leftrightarrow\dfrac{2x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}=0\\ \Leftrightarrow\dfrac{2x}{x-2}=0\\ \Rightarrow2x=0\\ \Leftrightarrow x=0\left(tm\right)\)

\(b,ĐKXĐ:x\ge\dfrac{5}{4}\\ \left|3x-2\right|=4x-5\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=4x-5\\3x-2=5-4x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\7x-7=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)