a: ĐKXĐ: y<>2; y<>-2
\(PT\Rightarrow\left(y-1\right)\left(y+2\right)-5\left(y-2\right)=12+y^2-4\)
=>y^2+y-2-5y+10=y^2+8
=>-4y+8=8
=>y=0(nhận)
b: ĐKXD: z<>2/3; z<>-2/3
\(PT\Leftrightarrow\dfrac{1}{\left(2z-3\right)^2}+\dfrac{3}{\left(2z-3\right)\left(2z+3\right)}=\dfrac{4}{\left(2z+3\right)^2}\)
=>\(\left(2z+3\right)^2+3\left(2z-3\right)\left(2z+3\right)=4\left(2z-3\right)^2\)
=>\(4z^2+12z+9+3\left(4z^2-9\right)=4\left(4z^2-12z+9\right)\)
=>\(16z^2+12z-18=16z^2-48z+36\)
=>60z=54
=>z=9/10(nhận)
d: ĐKXĐ: x<>1; x<>3
\(PT\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)
=>x^2+2x-15=x^2-9
=>2x-15=-9
=>x=3(loại)
c:
ĐKXĐ: u<>1/3; u<>-11/3
\(PT\Leftrightarrow\dfrac{-2}{\left(3u-1\right)\left(3u+11\right)}=\dfrac{1}{\left(3u-1\right)^2}-\dfrac{3}{\left(3u+11\right)^2}\)
=>\(-2\left(3u-1\right)\left(3u+11\right)=\left(3u+11\right)^2-3\left(3u-1\right)^2\)
=>\(-2\left(9u^2+30u-11\right)=9u^2+66u+121-3\left(9u^2-6u+1\right)\)
=>\(-18u^2-60u+22=-18u^2+84u+118\)
=>-60u+22=84u+118
=>-144u=96
=>u=-2/3(nhận)
