a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\)                               ĐKXĐ : x #0, x#2, x#-2

<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)

<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)

=> 10 - 2x + 7x - 14 = 4x - 4 + x

<=>-2x + 7x - 4x + x  = -4 - 10 + 14

<=>x=-14

\(ĐKXĐ:x\ne49;x\ne50\)

Đặt \(x-49=u;x-50=v\)

Phương trình trở thành \(\frac{50}{u}+\frac{49}{v}=\frac{u}{50}+\frac{v}{49}\)

\(\Rightarrow\frac{50v+49u}{uv}=\frac{49u+50v}{2450}\)

\(\Rightarrow\orbr{\begin{cases}50v+49u=0\\uv=2450\end{cases}}\)

+) \(50v+49u=0\)

\(\Rightarrow50v=-49u\)

\(\Rightarrow\frac{v}{-49}=\frac{u}{50}=\frac{\left(x-50\right)-\left(x-49\right)}{-49-50}\)

\(=\frac{-1}{-99}=\frac{1}{99}\)

\(\Rightarrow\hept{\begin{cases}v=\frac{-49}{99}\\u=\frac{50}{99}\end{cases}}\Rightarrow x=\frac{4901}{99}\)(tm)

+) \(uv=2450\)

hay \(\left(x-49\right)\left(x-50\right)=2450\)

\(\Leftrightarrow x^2-99x+2450=2450\)

\(\Leftrightarrow x^2-99x=0\Leftrightarrow x\left(x-99\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=99\end{cases}}\left(tm\right)\)

Vậy phương trình có 3 nghiệm \(S=\left\{0;\frac{4901}{99};99\right\}\)

ok cảm ơn bn

\(\frac{x+99}{-1}=\frac{y-98}{2}=\frac{z+97}{-3}=\frac{x+99-\left(y-98\right)+\left(z+97\right)}{-1-2+\left(-3\right)}=\frac{\left(x-y+z\right)+294}{-6}=\frac{50+294}{-6}=-\frac{172}{3}\)

x + 99 = 172/3  => x =-125/3

y - 98 = - 344/3 => y =  - 50 /3

z+ 97 = 172 => z = 75

\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{97}+\frac{x+100}{96}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

Dễ thấy \(\left(\frac{1}{99}< \frac{1}{98}< \frac{1}{97}< \frac{1}{96}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)\ne0\)

\(\Rightarrow x+100=0\Rightarrow x=-100\)

Vậy x = -100

\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}+4=0\)

\(\Rightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)

\(\Rightarrow\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)

\(\Rightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)=0\)

Dễ thấy \(\left(\frac{1}{91}>\frac{1}{93}>\frac{1}{95}>\frac{1}{97}\right)\)nên \(\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)\ne0\)

\(\Rightarrow200-x=0\Rightarrow x=200\)

Vậy x = 200

Đặt \(A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

=>  \(\frac{1}{5}.A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}+\frac{1}{5^{100}}\)

=> \(A-\frac{1}{5}A=\frac{4}{5}.A=1-\frac{1}{5^{100}}\Rightarrow\frac{4}{5}.A=\frac{5^{100}-1}{5^{100}}\Rightarrow A=\frac{5^{100}-1}{4.5^{99}}\)

Tính \(\frac{1}{50}+\frac{1}{150}+\frac{1}{300}+...+\frac{1}{9500}=\frac{1}{25}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{380}\right)\)

\(=\frac{1}{25}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right)=\frac{1}{25}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)\(=\frac{1}{25}.\left(1-\frac{1}{20}\right)=\frac{19}{20.25}=\frac{19}{4.5^3}\)

vậy phương trình đã cho trở thành:

\(\frac{5^{100}-1}{4.5^{99}}.x+\frac{1}{4.5^{99}.x}=\frac{19}{4.5^3}\Rightarrow\left(5^{100}-1\right)x^2+1=19.5^{96}.x\)

\(\left(5^{100}-1\right)x^2-19.5^{96}.x+1=0\)

bạn kiểm tra lại đề lần nữa, phương trình này có nghiệm  rất lẻ , nghiệm lớn