ĐKXĐ : \(x\ne0;x\ne\pm5\)

\(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)

\(\Leftrightarrow\frac{x+5}{x\left(x-5\right)}-\frac{x-5}{2x\left(x+5\right)}=\frac{x+25}{2\left(x-5\right)\left(x+5\right)}\)

\(\Leftrightarrow\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\frac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)

\(\Rightarrow2\left(x+5\right)^2-\left(x-5\right)^2=x\left(x+25\right)\)

\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)

\(\Leftrightarrow5x+25=0\)

\(\Leftrightarrow x=-5\)(ko t/m ĐKXĐ)

Vậy phương trình vô nghiệm.

errrrr

\(\frac{x+5}{x^2-5x}-\frac{x+25}{2x^2-50}=\frac{x-5}{2x^2+10x}\)

\(\Leftrightarrow\frac{x+5}{x\left(x-5\right)}-\frac{x+25}{2\left(x^2-25\right)}=\frac{x-5}{2x\left(x+5\right)}\)

\(\Leftrightarrow\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\frac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}=\frac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}\)

\(\Rightarrow2\left(x^2+10x+25\right)-x^2-25x=x^2-10x+25\)

\(\Leftrightarrow2x^2+20x+50-x^2-25x-x^2+10x-25=0\)

\(\Leftrightarrow5x+25=0\)

\(\Leftrightarrow x=-5\)

\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\)

\(\Leftrightarrow\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=\frac{16}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=16\)

\(\Leftrightarrow2.2x=16\)

\(\Leftrightarrow4x=16\)

\(\Leftrightarrow x=4\)

a)  \(\frac{x+5}{x^2-5x}-\frac{x+25}{2x^2-50}=\frac{x-5}{2x^2+10x}\)
\(\Leftrightarrow\frac{x+5}{x\left(x-5\right)}-\frac{x+25}{2\left(x^2-25\right)}=\frac{x-5}{2x\left(x+5\right)}\)

\(ĐKXĐ:x\ne0;x\ne\pm5\)

\(MTC:2x\left(x^2-25\right)=2x\left(x-5\right)\left(x+5\right)\)

\(\Leftrightarrow\frac{2\left(x+5\right)\left(x+5\right)}{2x\left(x^2-25\right)}-\frac{x\left(x+25\right)}{2x\left(x^2-25\right)}=\frac{\left(x-5\right)\left(x-5\right)}{2x\left(x^2-25\right)}\)

\(\Rightarrow2\left(x+5\right)\left(x+5\right)-x\left(x+25\right)=\left(x-5\right)\left(x-5\right)\)

\(\Leftrightarrow2\left(x+5\right)^2-x\left(x+25\right)=\left(x-5\right)^2\)

\(\Leftrightarrow2\left(x^2+10x+25\right)-\left(x^2+25x\right)=x^2-10x+25\)

\(\Leftrightarrow2x^2+20x+50-x^2-25x=x^2-10x+25\)

\(\Leftrightarrow x^2-5x+50=x^2-10x+25\)

\(\Leftrightarrow x^2-5x+50-x^2+10x-25=0\)

\(\Leftrightarrow5x+25=0\)

\(\Leftrightarrow5\left(x+5\right)=0\)

\(\Leftrightarrow x+5=0\)

\(\Leftrightarrow x=-5\left(L\right)\)

Vậy \(S=\varnothing\)

b)  \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\)

\(ĐKXĐ:x\ne\pm1\)

\(MTC:\left(x-1\right)\left(x+1\right)=x^2-1\)

\(\Leftrightarrow\frac{\left(x+1\right)^2}{x^2-1}-\frac{\left(x-1\right)^2}{x^2-1}=\frac{16}{x^2-1}\)

\(\Rightarrow\left(x+1\right)^2-\left(x-1\right)^2=16\)

\(\Leftrightarrow\left(x+1+x-1\right)\left(x+1-x+1\right)=16\)

\(\Leftrightarrow4.x=16\)

\(\Leftrightarrow x=4\left(N\right)\)

Vậy \(S=\left\{4\right\}\)

ĐK: ...

c) \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)

\(\Leftrightarrow\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=\frac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)

\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)

\(\Leftrightarrow5x+25=0\)

\(\Leftrightarrow x=-5\)( ko t/m )

d) tương tự, ngại tính lắm

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

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

\(\Leftrightarrow4x^2-3x-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=\frac{-1}{4}\left(c\right)\end{matrix}\right.\)