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1) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)
\(\Leftrightarrow-23x-7=0\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)
Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)
2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)
Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
Suy ra: \(9x^2+6x+24x+16=9x^2\)
\(\Leftrightarrow30x+16=0\)
\(\Leftrightarrow30x=-16\)
hay \(x=-\dfrac{8}{15}\)(nhận)
Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)
e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)
\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)
\(\Leftrightarrow x=-1\left(TM\right)\)
a: Ta có: \(6-4x=5(x+3)+3\)
\(\Leftrightarrow6-4x-5x-12-3=0\)
\(\Leftrightarrow-9x=9\)
hay x=-1
b: Ta có: \(\dfrac{x+3}{2}-1=\dfrac{x-1}{3}+\dfrac{x+5}{6}\)
\(\Leftrightarrow15x+45-30=10x-30+5x+25\)
\(\Leftrightarrow15=-5\left(loại\right)\)
c: Ta có: \(\left(x-2\right)\left(2x+1\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow2\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
d: Ta có: \(\dfrac{2}{x^2-2x}+\dfrac{1}{x}=\dfrac{x+2}{x-2}\)
\(\Leftrightarrow2+x-2=x^2+2x\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
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) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)
Suy ra: x+2=0
hay x=-2(thỏa ĐK)
Vậy: S={-2}
d)
ĐKXĐ: \(x\notin\left\{1;3\right\}\)
Ta có: \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)
\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)
Suy ra: \(x^2-3x+5x-15=x^2-1-8\)
\(\Leftrightarrow2x-15+9=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3(loại)
Vậy: \(S=\varnothing\)
d: Ta có: \(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)
\(\Leftrightarrow x^2-3x-2x^2-6x+3x=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow-x\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)
\(\dfrac{9x-2}{x^2-x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\left(x\ne-2;x\ne3\right)\\ < =>\dfrac{9x-2}{x^2-3x+2x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\\ < =>\dfrac{9x-2}{x\left(x-3\right)+2\left(x-3\right)}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\\ < =>\dfrac{9x-2}{\left(x-3\right)\left(x+2\right)}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\)
suy ra: \(9x-2+2x\left(x-3\right)-\left(x-1\right)\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)
\(< =>9x-2+2x^2-6x-\left(x^2+2x-x-2\right)=x^2+2x-3x-6\)
\(< =>9x-2+2x^2-6x-x^2-2x+x+2=x^2-x-6\)
\(< =>2x^2-x^2-x^2+9x-6x-2x+x+x=6+2-2\)
\(< =>3x=6\\ < =>x=2\left(tm\right)\)
ĐKXĐ: \(x\ne\left\{-2;3\right\}\)
\(\dfrac{9x-2}{x^2-x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\)
\(\Leftrightarrow\dfrac{9x-2}{\left(x+2\right)\left(x-3\right)}+\dfrac{2x\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow9x-2+2x\left(x-3\right)-\left(x-1\right)\left(x+2\right)=\left(x+2\right)\left(x-3\right)\)
\(\Leftrightarrow9x-2+2x^2-6x-x^2-x+2=x^2-x-6\)
\(\Leftrightarrow3x=-6\)
\(\Leftrightarrow x=-2\left(loại\right)\)
Vậy: PT vô nghiệm.