Bài 1:
a) ĐKXĐ: \(x\notin\left\{1;-3\right\}\)
Ta có: \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{\left(x-1\right)\left(x+3\right)}=1\)
\(\Leftrightarrow\dfrac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}+\dfrac{4}{\left(x-1\right)\left(x+3\right)}=\dfrac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}\)
Suy ra: \(3x^2+9x-x-3-\left(2x^2-2x+5x-5\right)+4=x^2+3x-x-3\)
\(\Leftrightarrow3x^2+8x-3-2x^2-3x+5+4=x^2+2x-3\)
\(\Leftrightarrow x^2+5x+6-x^2-2x+3=0\)
\(\Leftrightarrow3x+9=0\)
\(\Leftrightarrow3x=-9\)
hay x=-3(loại)
Vậy: \(S=\varnothing\)
b) ĐKXĐ: \(x\notin\left\{-3;1\right\}\)
Ta có: \(\dfrac{x+2}{x+3}-\dfrac{x+1}{x-1}=\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
Suy ra: \(x^2-x+2x-2-\left(x^2+4x+3\right)=4\)
\(\Leftrightarrow x^2+x-2-x^2-4x-3-4=0\)
\(\Leftrightarrow-3x-9=0\)
\(\Leftrightarrow-3x=9\)
hay x=-3(loại)
Vậy: \(S=\varnothing\)
c) ĐKXĐ: \(x\notin\left\{2;5\right\}\)
Ta có: \(\dfrac{3x}{x-2}-\dfrac{x}{x-5}+\dfrac{2x^2}{7x-10-x^2}=0\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x-5\right)}=0\)
Suy ra: \(3x^2-15x-x^2+2x-2x^2=0\)
\(\Leftrightarrow-13x=0\)
hay x=0(nhận)
Vậy: S={0}
d) ĐKXĐ: \(x\notin\left\{5;-5;25\right\}\)
Ta có: \(\dfrac{3}{4x-20}+\dfrac{15}{50-2x}+\dfrac{7}{6x+30}=0\)
\(\Leftrightarrow\dfrac{3}{4\left(x-5\right)}-\dfrac{15}{2\left(x-25\right)}+\dfrac{7}{6\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{18\left(x+5\right)}{24\left(x-5\right)\left(x+5\right)}=\dfrac{15}{2\left(x-25\right)}\)
\(\Leftrightarrow36\left(x+5\right)\left(x-25\right)=360\left(x^2-25\right)\)
\(\Leftrightarrow36\left(x^2-25x+5x-125\right)=360x^2-9000\)
\(\Leftrightarrow36x^2-720x-4500-360x^2+9000=0\)
\(\Leftrightarrow-324x^2-720x+4500=0\)
\(\Leftrightarrow-18\left(18x^2+40x-250\right)=0\)
\(\Leftrightarrow18x^2+40x-250=0\)
\(\Leftrightarrow18x^2+90x-50x-250=0\)
\(\Leftrightarrow18x\left(x+5\right)-50\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(18x-50\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\18x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\left(loại\right)\\18x=50\end{matrix}\right.\Leftrightarrow x=\dfrac{25}{9}\)
Vậy: \(S=\left\{\dfrac{25}{9}\right\}\)
