ĐKXĐ : \(x\ne\left\{5;-5;0\right\}\)

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

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

=> \(2x-x+5=4x+20\)

<=> \(4x-2x+x=5-20\)

<=> \(3x=-15\) <=> \(x=-5\) ( ko t/m )

Vậy pt vô nghiệm.

ĐKXĐ : x khác 0; x khác 5 ; x khác -5

\(\frac{2}{x^2-25}+\frac{1}{x^2+5x}=\frac{4}{x\left(x-5\right)}\Leftrightarrow\frac{2x}{x\left(x-5\right)\left(x+5\right)}+\frac{x-5}{x\left(x-5\right)\left(x+5\right)}=\frac{4\left(x+5\right)}{x\left(x-5\right)\left(x+5\right)}\)

\(\Leftrightarrow\frac{2x+x-5}{x\left(x-5\right)\left(x+5\right)}=\frac{4\left(x+5\right)}{x\left(x-5\right)\left(x+5\right)}\Rightarrow3x-5=4x+20\)

\(\Leftrightarrow3x-4x=20+5\Leftrightarrow-x=25\Leftrightarrow x=-25\)( thỏa mãn ĐKXĐ)

Vậy phương trình có nghiệm x = -25

\(\frac{2}{x^2-25}-\frac{1}{x^2+5x}=\frac{4}{x\left(x-5\right)}\)

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

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

<=> \(2x-\left(x-5\right)=4\left(x+5\right)\)

<=> \(2x-x+5=4x+20\)

<=> \(x+5=4x+20\)

<=> \(x-4x=20-5\)

<=> \(-3x=15< =>x=-5\)

Vậy ....

ĐKXĐ:...

\(\left(\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\frac{25-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}+\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)

\(=\left(\frac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}\right):\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)=\frac{-5}{\left(\sqrt{x}+5\right)}.\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}{\left(9-x\right)}\)

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

\(a,⇔\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)

\(⇔(x-23)(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27})=0\)

\(⇔x-23=0\) (vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\))

\(⇔x=23\)

\(b,⇔\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}+\frac{x+100}{95}=0\)

\(⇔(x+100)(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95})=0\)

\(⇔x+100=0\) (vì \(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}>0\))

\(⇔x=-100\)

\(c,⇔(\frac{x+1}{2012}+1)+(\frac{x+2}{2011}+1)=(\frac{x+3}{2010}+1)+(\frac{x+4}{2009}+1)\)

\(⇔\frac{x+2013}{2012}+\frac{x+2013}{2011}-\frac{x+2013}{2010}-\frac{x+2013}{2009}=0\)

\(⇔(x+2013)(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009})=0\)

\(⇔x+2013=0\) (vì \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}<0\))

\(⇔x=-2013\)

ĐKXĐ :\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3\ne0\\\sqrt{x}+5\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne3\\\sqrt{x}\ne-5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

- Ta có : \(\left(\frac{x-5\sqrt{x}}{25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{x-9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\frac{x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{-x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right):\left(\frac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\frac{x-5\sqrt{x}-25}{25}\right)\left(\frac{\sqrt{x}+5}{-\sqrt{x}-3}\right)\)

\(=\frac{\left(x-5\sqrt{x}-25\right)\left(\sqrt{x}+5\right)}{-25\left(\sqrt{x}+3\right)}=\frac{x\sqrt{x}+5x-5x-25\sqrt{x}-25\sqrt{x}-125}{-25\left(\sqrt{x}+3\right)}\)

\(=\frac{x\sqrt{x}-125-50\sqrt{x}}{-25\left(\sqrt{x}+3\right)}\)