1) \(\dfrac{2x+3}{2x-3}-\dfrac{3}{4x-6}=\dfrac{2}{5}\left(x\ne\dfrac{3}{2}\right)\)
\(\Leftrightarrow\dfrac{2x+3}{2x-3}-\dfrac{3}{2\left(2x-3\right)}=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{10\left(2x+3\right)}{10\left(2x-3\right)}-\dfrac{15}{10\left(2x-3\right)}=\dfrac{4\left(2x-3\right)}{10\left(2x-3\right)}\)
\(\Leftrightarrow20x+30-15=8x-12\)
\(\Leftrightarrow20x-8x=-12-30+15\)
\(\Leftrightarrow12x=-27\)
\(\Leftrightarrow x=-\dfrac{27}{12}=-\dfrac{9}{4}\left(thỏa.đk\right)\)
2) \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\left(x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow x^2+3x+2-\left(5x-10\right)=12+x^2-4\)
\(\Leftrightarrow x^2-x^2+3x+2-5x+10=12-4\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\left(loại.ko.thỏa.đk\right)\)
Vậy phương trình đã cho vô nghiệm.
4: ĐKXĐ: x<>1
\(\dfrac{1}{x-1}+\dfrac{2x^2-5}{x^3-1}=\dfrac{4}{x^2+x+1}\)
=>\(\dfrac{x^2+x+1+2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4x-4}{\left(x-1\right)\left(x^2+x+1\right)}\)
=>3x^2+x-4=4x-4
=>3x^2-3x=0
=>x(3x-3)=0
=>x=0(nhận) hoặc x=1(loại)
5: ĐKXĐ: x<>1; x<>-1
\(PT\Leftrightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=3x\cdot\dfrac{x+1-x+1}{x+1}\)
\(\Leftrightarrow\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{6x}{x+1}\)
=>\(\dfrac{4x}{\left(x-1\right)\left(x+1\right)}=\dfrac{6x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
=>6x(x-1)=4x
=>6x^2-6x-4x=0
=>6x^2-10x=0
=>3x^2-5x=0
=>x(3x-5)=0
=>x=0 hoặc x=5/3
6: ĐKXĐ: x<>-3; x<>1
\(PT\Leftrightarrow\dfrac{2x}{x-1}+\dfrac{4}{\left(x+3\right)\left(x-1\right)}=\dfrac{2x-5}{x+3}\)
=>\(\dfrac{2x\left(x+3\right)+4}{\left(x+3\right)\left(x-1\right)}=\dfrac{\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)
=>2x^2+6x+4=2x^2-2x-5x+5
=>6x+4=-7x+5
=>13x=1
=>x=1/13(nhận)
7: ĐKXĐ: x<>4; x<>2
\(PT\Leftrightarrow\dfrac{x+3}{x-4}+\dfrac{x-1}{x-2}=\dfrac{-2}{\left(x-2\right)\left(x-4\right)}\)
=>\(\dfrac{\left(x+3\right)\left(x-2\right)+\left(x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x-2\right)}=-\dfrac{2}{\left(x-2\right)\left(x-4\right)}\)
=>x^2+x-6+x^2-5x+4=-2
=>2x^2-4x=0
=>2x(x-2)=0
=>x=2(loại) hoặc x=0(nhận)
8: ĐKXĐ: x<>1; x<>-3
\(PT\Leftrightarrow\dfrac{\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+2x-3-4}{\left(x+3\right)\left(x-1\right)}\)
=>3x^2+9x-x-3-2x^2+2x-5x+5=x^2+2x-7
=>x^2+5x+2=x^2+2x-7
=>3x=-9
=>x=-3(loại)
3) \(\dfrac{2}{x^2+2x+1}-\dfrac{5}{x^2-2x+1}=\dfrac{3}{1-x^2}\left(x\ne\pm1\right)\)
\(\Leftrightarrow\dfrac{2}{\left(x+1\right)^2}-\dfrac{5}{\left(x-1\right)^2}=-\dfrac{3}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow\dfrac{2\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}-\dfrac{5\left(x+1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}=-\dfrac{3\left(x^2-1\right)}{\left(x-1\right)^2\left(x+1\right)^2}\)
\(\Leftrightarrow2\left(x^2-2x+1\right)-5\left(x^2+2x+1\right)=-3\left(x^2-1\right)\)
\(\Leftrightarrow2x^2-4x+2-5x^2-10x-5=-3x^2+3\)
\(\Leftrightarrow16x=-6\)
\(\Leftrightarrow x=-\dfrac{6}{16}=-\dfrac{3}{8}\left(thỏa.đkxđ\right)\)
