a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x\left(1-\dfrac{x-1}{x+1}\right)\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=3x\cdot\dfrac{x+1-x+1}{x+1}\)
\(\Leftrightarrow\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{6x}{x+1}\)
\(\Leftrightarrow\dfrac{4x}{\left(x-1\right)\left(x+1\right)}=\dfrac{6x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
Suy ra: \(4x=6x\left(x-1\right)\)
\(\Leftrightarrow4x-6x^2+6x=0\)
\(\Leftrightarrow-6x^2+10x=0\)
\(\Leftrightarrow x\left(-6x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=\dfrac{5}{3}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{0;\dfrac{5}{3}\right\}\)
b)
ĐKXĐ: \(x\notin\left\{-3;1\right\}\)
Ta có: \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)
\(\Leftrightarrow\dfrac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}+\dfrac{4}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+2x-3}{\left(x+3\right)\left(x-1\right)}\)
Suy ra: \(3x^2+9x-x-3-\left(2x^2-2x+5x-5\right)+4=x^2+2x-3\)
\(\Leftrightarrow3x^2+8x-3-2x^2-3x+5+4-x^2-2x+3=0\)
\(\Leftrightarrow3x+9=0\)
\(\Leftrightarrow3x=-9\)
hay x=-3(loại)
Vậy: \(S=\varnothing\)
c) Ta có: \(\dfrac{2}{x-3}=\dfrac{2x+5}{x^2-x}\)
\(\Leftrightarrow\dfrac{2\left(x^2-x\right)}{x\left(x-1\right)\left(x-3\right)}=\dfrac{\left(2x+5\right)\left(x-3\right)}{x\left(x-1\right)\left(x-3\right)}\)
Suy ra: \(2x^2-2x=2x^2-6x+5x-15\)
\(\Leftrightarrow-2x+6x-5x=-15\)
\(\Leftrightarrow-x=-15\)
hay x=15(thỏa ĐK)
Vậy: S={15}
d)
ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{1}{2x-2}-\dfrac{2x+1}{x^2+x+1}+\dfrac{3}{2x+2}=0\)
\(\Leftrightarrow\dfrac{x+1}{2\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{2x+1}{x^2+x+1}=0\)
\(\Leftrightarrow\dfrac{x+1+3x-3}{2\left(x-1\right)\left(x+1\right)}-\dfrac{2x+1}{x^2+x+1}=0\)
\(\Leftrightarrow\dfrac{4x-2}{2\left(x-1\right)\left(x+1\right)}-\dfrac{2x+1}{x^2+x+1}=0\)
\(\Leftrightarrow\dfrac{2x-1}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x+1}{x^2+x+1}=0\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}-\dfrac{\left(2x+1\right)\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}=0\)
Suy ra: \(2x^3+2x^2+2x-x^2-x-1-\left(2x^3-2x+x^2-1\right)=0\)
\(\Leftrightarrow2x^3+x^2+x-1-2x^3+2x-x^2+1=0\)
\(\Leftrightarrow3x=0\)
hay x=0
Vậy:S={0}
e)
ĐKXĐ: \(x\notin\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)
Ta có: \(\dfrac{12x+1}{6x-2}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(9x^2-1\right)}\)
\(\Leftrightarrow\dfrac{2\left(12x+1\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}-\dfrac{4\left(9x-5\right)\left(3x-1\right)}{4\left(3x-1\right)\left(3x+1\right)}=\dfrac{108x-36x^2-9}{4\left(3x-1\right)\left(3x+1\right)}\)
Suy ra: \(2\left(36x^2+12x+3x+1\right)-4\left(27x^2-9x-15x+5\right)=108x-36x^2-9\)
\(\Leftrightarrow72x^2+30x+2-108x^2+96x-20-108x+36x^2+9=0\)
\(\Leftrightarrow18x-9=0\)
\(\Leftrightarrow18x=9\)
hay \(x=\dfrac{1}{2}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
