k: ĐKXĐ: x<>5; x<>-5

\(\frac{7}{6x+30}+\frac{3}{4x-20}=\frac{15}{2x^2-50}\)

=>\(\frac{7}{6\left(x+5\right)}+\frac{3}{4\left(x-5\right)}=\frac{15}{2\left(x-5\right)\left(x+5\right)}\)

=>\(\frac{7\cdot2\cdot\left(x-5\right)}{12\left(x-5\right)\left(x+5\right)}+\frac{3\cdot3\cdot\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}=\frac{90}{12\left(x-5\right)\left(x+5\right)}\)

=>14(x-5)+9(x+5)=90

=>14x-70+9x+45=90

=>23x-25=90

=>23x=115

=>x=5(loại)

l: ĐKXĐ: x<>1; x<>-1

\(\frac{x}{x-1}+\frac{2x}{1-x^2}=0\)

=>\(\frac{x}{x-1}-\frac{2x}{\left(x-1\right)\left(x+1\right)}=0\)

=>\(\frac{x\left(x+1\right)-2x}{\left(x-1\right)\left(x+1\right)}=0\)

=>\(\frac{x\left(x+1-2\right)}{\left(x-1\right)\left(x+1\right)}=0\)

=>\(\frac{x}{x+1}=0\)

=>x=0(nhận)

m: ĐKXĐ: x<>2; x<>-2

\(\frac{x}{x+2}+\frac{5x+3}{x^2-4}=\frac{x}{x-2}\)

=>\(\frac{x\left(x-2\right)+5x+3}{\left(x-2\right)\left(x+2\right)}=\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

=>x(x-2)+5x+3=x(x+2)

=>\(x^2-2x+5x+3=x^2+2x\)

=>3x+3=2x

=>3x-2x=-3

=>x=-3(nhận)

n: ĐKXĐ: x<>8

\(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac18=\frac{13x-102}{3x-24}\)

=>\(\frac{3\cdot12}{24\left(x-8\right)}+\frac{24\left(3x-20\right)}{24\left(x-8\right)}+\frac{3\left(x-8\right)}{24\left(x-8\right)}=\frac{8\left(13x-102\right)}{24\left(x-8\right)}\)

=>36+72x-480+3x-24=104x-816

=>104x-816=75x-468

=>29x=-468+816=348

=>x=12(nhận)

o: ĐKXĐ: x<>1/2; x<>-1/2

\(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}=\frac{8}{4x^2-1}\)

=>\(\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}=\frac{8}{\left(2x-1\right)\left(2x+1\right)}\)

=>\(\left(2x+1\right)^2-\left(2x-1\right)^2=8\)

=>\(4x^2+4x+1-\left(4x^2-4x+1\right)=8\)

=>8x=8

=>x=1(nhận)

p:

ĐKXĐ: x<>-1; x<>2

\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)

=>\(\frac{x-2-5\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\frac{-15}{\left(x+1\right)\left(x-2\right)}\)

=>x-2-5(x+1)=-15

=>x-2-5x-5=-15

=>-4x-7=-15

=>4x+7=15

=>4x=8

=>x=2(loại)

q: ĐKXĐ: x<>1; x<>3

\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)

=>\(\frac{\left(x+5\right)\cdot\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)

=>(x+5)(x-3)=(x+1)(x-1)-8

=>\(x^2+2x-15=x^2-1-8\)

=>2x-15=-9

=>2x=6

=>x=3(loại)

r: ĐKXĐ: x<>1/3; x<>-1/3

\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

=>\(\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)

=>\(\left(3x-1\right)^2-\left(3x+1\right)^2=12\)

=>\(9x^2-6x+1-9x^2-6x-1=12\)

=>-12x=12

=>x=-1(nhận)

s: ĐKXĐ: x<>2; x<>-2

\(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{x^2-4}+1\)

=>\(\frac{\left(x+1\right)\left(x+2\right)-5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{12+x^2-4}{\left(x-2\right)\left(x+2\right)}\)

=>(x+1)(x+2)-5(x-2)=\(x^2+8\)

=>\(x^2+3x+2-5x+10=x^2+8\)

=>-2x+12=8

=>-2x=-4

=>x=2(loại)