a: (x-2)(3x+5)=x2-4
=>(x-2)(3x+5)-(x-2)(x+2)=0
=>(x-2)(3x+5-x-2)=0
=>(x-2)(2x+3)=0
=>\(\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b: ĐKXĐ: x<>2
\(\dfrac{x^2+3x}{x^3-8}-\dfrac{4}{2-x}=\dfrac{5x-7}{x^2+2x+4}\)
=>\(\dfrac{x^2+3x}{\left(x-2\right)\left(x^2+2x+4\right)}+\dfrac{4\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{\left(5x-7\right)\left(x-2\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)
=>\(x^2+3x+4x^2+8x+16=5x^2-17x+14\)
=>11x+16=-17x+14
=>28x=-2
=>\(x=-\dfrac{1}{14}\left(nhận\right)\)
c: ĐKXĐ: x<>-2; y<>x/2
\(\left\{{}\begin{matrix}\dfrac{4}{x+2}-\dfrac{3}{x-2y}=\dfrac{11}{15}\\\dfrac{3}{x+2}+\dfrac{5}{x-2y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{20}{x+2}-\dfrac{15}{x-2y}=\dfrac{11}{3}\\\dfrac{9}{x+2}+\dfrac{15}{x-2y}=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{20}{x+2}-\dfrac{15}{x-2y}+\dfrac{9}{x+2}+\dfrac{15}{x-2y}=\dfrac{11}{3}+6=\dfrac{29}{3}\\\dfrac{3}{x+2}+\dfrac{5}{x-2y}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{29}{x+2}=\dfrac{29}{3}\\\dfrac{5}{x-2y}=2-\dfrac{3}{x+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2=3\\\dfrac{5}{x-2y}=2-\dfrac{3}{3}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3-2=1\\x-2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\2y=x-5=1-5=-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\left(nhận\right)\)
