a) \(\dfrac{x^2+5}{25-x^2}=\dfrac{3}{x+5}+\dfrac{x}{x-5}\)
\(\Leftrightarrow\dfrac{x^2+5}{5^2-x^2}=\dfrac{3\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow\dfrac{x^2+5}{5^2-x^2}=\dfrac{3\left(x-5\right)+x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow\dfrac{-\left(x^2+5\right)}{x^2-5^2}=\dfrac{3x-15+x^2+5x}{x^2-5^2}\)
\(\Leftrightarrow\dfrac{-\left(x^2+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{8x-15+x^2}{\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow-\left(x^2+5\right).\left(x-5\right)\left(x+5\right)=\left(x-5\right)\left(x+5\right)\left(8x-15+x^2\right)\)
\(\Leftrightarrow-\left(x^2+5\right)\left(x-5\right)\left(x+5\right)-\left(x-5\right)\left(x+5\right)\left(8x-15+x^2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(-x^2-5+8x-15+x^2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(-20+8x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+5=0\\-20x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\\x=\dfrac{2}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là S={5,-5,2/5}
a: \(\Leftrightarrow\dfrac{-x^2-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{3x-15+x^2+5x}{\left(x+5\right)\left(x-5\right)}\)
=>x^2+8x-15=-x^2-5
=>2x^2+8x-10=0
=>x^2+4x-5=0
=>(x+5)(x-1)=0
=>x=1(nhận) hoặc x=-5(loại)
b: \(\Leftrightarrow3x+6-2x-2=4x+5\)
=>4x+5=x+4
=>3x=-1
=>x=-1/3