\(\Leftrightarrow\frac{200\left(x+20\right)}{2x\left(x+20\right)}-\frac{240x}{2x\left(x+20\right)}=\frac{x\left(x+20\right)}{2x\left(x+20\right)}\) đk: x\(\ne0\) , x \(\ne-20\)

\(\Rightarrow200x+4000-240x=x^2+20x\)

\(\Leftrightarrow-x^2-60x+4000=0\)

\(\Leftrightarrow x^2+60x-4000=0\)

\(\Leftrightarrow x^2+100x-40x-4000=0\)

\(\Leftrightarrow\left(x+100\right)\left(x-40\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+100=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-100\left(tmđk\right)\\x=40\left(tmđk\right)\end{matrix}\right.\)

Vậy S\(=\left\{-100;40\right\}\)

\(\frac{100}{x}-\frac{120}{x+20}=\frac{1}{2}\)

\(\Leftrightarrow\frac{100}{x}-\frac{120}{x+20}=\frac{1}{2},x\ne0,x\ne-20\)

\(\Leftrightarrow\frac{100}{x}-\frac{120}{x+20}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{200\left(x+20\right)-240x-x\left(x+20\right)}{2x\left(x+20\right)}=0\)

\(\Leftrightarrow\frac{200x+4000-240x-x^2-20x}{2x\left(x+20\right)}=0\)

\(\Leftrightarrow-60x+4000-x^2=0\)

\(\Leftrightarrow-x^2-60x+4000=0\)

\(\Leftrightarrow x^2+60x-4000=0\)

\(\Leftrightarrow\frac{-60\pm\sqrt{60^2}-4.1\left(-4000\right)}{2}\)

\(\Leftrightarrow\frac{-60\pm\sqrt{3600+16000}}{2}\)

\(\Leftrightarrow\frac{-60\pm\sqrt{19600}}{2}\)

\(\Leftrightarrow\frac{-60\pm140}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{-60+140}{2}\\\frac{-60-140}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=40\\x=-100\end{matrix}\right.,x\ne0,x\ne-20\)

\(\frac{96}{x-4}-\frac{120}{x+4}=1\)

\(\Leftrightarrow\frac{96}{x-4}-\frac{120}{x+4}-1=0\)

\(\Leftrightarrow\frac{96\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{120\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\frac{\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=0\)

\(\Leftrightarrow\frac{96x+384}{x^2-2^2}-\frac{120x-480}{x^2-2^2}-\frac{x^2-2^2}{x^2-2^2}=0\)

\(\Leftrightarrow\left(96x+384-120x+480-x^2+2^2\right)\cdot\frac{1}{x^2-2^2}=0\)

\(\Leftrightarrow-x^2-24x+868=0\)

\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)

\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)

\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)

\(\Leftrightarrow x+116=0\Leftrightarrow x=-116\)

\(\frac{x-1}{117}+\frac{x-2}{118}+\frac{x-3}{119}=\frac{x-4}{120}+\frac{x-5}{121}+\frac{x-6}{122}\)

\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}+1=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)

\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)

\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)

Vì \(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\ne0\)

Nên x + 116 = 0

<=> x = -116

1/Tôi chỉ bt 1 câu thui thông cảm :)

P=\(\frac{x}{x-1}+\frac{4}{x+1}+\frac{4-6x}{x^2-1}\)       ĐK:\(\hept{\begin{cases}x-1\ne0\\x+1\ne\\x^2-1\ne0\end{cases}1}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-1\\x\ne1\end{cases}}\)

P=\(\frac{x\left(x+1\right)+4\left(x-1\right)+4-6x}{\left(x-1\right).\left(x+1\right)}\) 

=\(\frac{x^2+x+4x-4+4-6x}{\left(x-1\right)\left(x+1\right)}=\frac{x^2-x}{\left(x-1\right)\left(x+1\right)}\)

=\(\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x}{x+1}\)

^^ học tốt!

1/

\(đkxđ\Leftrightarrow x\ne\pm1\)

\(P=\frac{x}{x-1}+\frac{4}{x+1}+\frac{4-6x}{x^2-1}\)

\(=\frac{x}{x-1}+\frac{4}{x+1}+\frac{4-6x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{4\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{4-6x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2+x+4x-4+4-6x}{\left(x-1\right)\left(x+1\right)}\)

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

2/

Kẻ \(DE//AB\left(E\in AC\right)\)

\(\Rightarrow\frac{DE}{AB}=\frac{EC}{AC}\)

\(\Delta ADE\)đều (vì .............)\(\Rightarrow AD=AE=DE\)

\(\Rightarrow\frac{AD}{AB}=\frac{AC-AE}{AC}\)mà \(AE=AD\)

\(\Rightarrow\frac{AB}{AB}=1-\frac{AD}{AC}\)

\(\Rightarrow\frac{AD}{AB}+\frac{AD}{AC}=1\)

\(\Rightarrow AD\left(\frac{1}{AB}+\frac{1}{AC}\right)=1\)

\(\Rightarrow\frac{1}{AB}+\frac{1}{AC}=\frac{1}{AD}\left(ĐPCM\right)\)