Sửa đề:

 \(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

ĐKXĐ: \(x\notin\left\{1;3;8;20\right\}\)

PT=>\(-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

=>\(-\dfrac{1}{x-4}=-\dfrac{3}{4}\)

=>\(x-1=\dfrac{4}{3}\)

=>\(x=\dfrac{4}{3}+1=\dfrac{7}{3}\)(nhận)

\(ĐKXĐ:x\ne1,x\ne3,x\ne8,x\ne20\)

\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{2\left(x-8\right)\cdot\left(x-20\right)+5\left(x-1\right)\cdot\left(x-20\right)+12\left(x-1\right)\cdot\left(x-3\right)-\left(x-1\right)\cdot\left(x-3\right)\cdot\left(x-8\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{\left(2x-16\right)\cdot\left(x-20\right)+\left(5x-5\right)\cdot\left(x-20\right)+\left(12x-12\right)\cdot\left(x-3\right)-\left(x^2-3x-x+3\right)\cdot\left(x-8\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-12x+36-\left(x^2-4x+3\right)\cdot\left(x-8\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-\left(x^3-8x^2-4x^2+32x+3x-24\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}\)

\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-12x+36-\left(x^3-12x^2+35x-24\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-12x+36-x^3+12x^2-35x+24}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{31x^2-244x+480-x^3}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-x^3+31x^2-244x+480}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-x^3+3x^2+28x^2-84x-160x+480}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-x^2\cdot\left(x-3\right)+28x\cdot\left(x-3\right)-160\left(x-3\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-\left(x-3\right)\left(x^2-28x+160\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-1\left(x^2-8x-20x+160\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-1\left(x^2-8x-20x+160\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-1\left(x\cdot\left(x-8\right)-20\left(x-8\right)\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-1\left(x-20\right)\left(x-8\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-1}{x-1}=-\dfrac{3}{4}\)

\(\Leftrightarrow-\dfrac{1}{x-1}=-\dfrac{3}{4}\)

\(\Leftrightarrow-4=-3\left(x-1\right)\)

\(\Leftrightarrow-4=-3\left(x-1\right)\)

\(\Leftrightarrow-4=-3x+3\)

\(\Leftrightarrow3x=3+4\)

\(\Leftrightarrow3x=7\)

\(\Rightarrow x=\dfrac{7}{3}\)

Vậy \(x=\dfrac{7}{3}\)

\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}+\dfrac{1}{x-20}=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}+\dfrac{1}{x-20}=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{x-1}=\dfrac{3}{4}\Rightarrow3x-3=4\Rightarrow x=\dfrac{7}{3}\)

Vậy...

\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-1}{4}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-1}{4}\)

\(\Leftrightarrow\dfrac{1}{x-1}=\dfrac{-1}{4}\)

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

\(\Leftrightarrow x-1=-4\)

\(\Leftrightarrow x=-3\left(tm\right)\)

Vậy ..............

\(\Leftrightarrow\dfrac{2}{x-3}-\dfrac{2}{x-2}+\dfrac{1}{x-8}-\dfrac{1}{x-3}+\dfrac{1}{x-20}-\dfrac{1}{x-8}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)

\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{2}{x-2}=\dfrac{-3}{4}\)

\(\Leftrightarrow4\left(x-2\right)-8\left(x-3\right)=-3\left(x-3\right)\left(x-2\right)\)

\(\Leftrightarrow4x-8-8x+24+3\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow3x^2-15x+18-4x+16=0\)

\(\Leftrightarrow3x^2-19x+34=0\)

\(\text{Δ}=\left(-19\right)^2-4\cdot3\cdot34=-47< 0\)

Do đó: Phương trình vô nghiệm

\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}+\dfrac{1}{x-20}=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{x}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}+\dfrac{1}{x-20}=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{x-1}=\dfrac{3}{4}\Rightarrow3x-3=4\Rightarrow x=\dfrac{7}{3}\)

Chúc bạn học tốt!

a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)

=>x+1=0

hay x=-1

b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)

=>x-2010=0

hay x=2010

c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)

=>x=15

\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}=\dfrac{-3}{4}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}+\dfrac{1}{x-20}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

Đến đây cạn rồi?! ==''

a) Ta có: \(C=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^3}{1-x}+x\right)\left(\dfrac{1+x^3}{1+x}-x\right)\right]\)

\(=\dfrac{x\left(x^2-1\right)^2}{x^2+1}:\left[\left(\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}+x\right)\left(\dfrac{\left(1+x\right)\left(1-x+x^2\right)}{\left(1+x\right)}-x\right)\right]\)

\(=\dfrac{x\left(x^2-1\right)^2}{x^2+1}:\left[\left(x^2+2x+1\right)\left(x^2-2x+1\right)\right]\)

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

\(=\dfrac{x}{x^2+1}\)

b) Thay \(x=-\dfrac{3}{2}\) vào C, ta được:

\(C=\dfrac{-3}{2}:\left(\dfrac{9}{4}+1\right)=\dfrac{-3}{2}:\dfrac{13}{4}=\dfrac{-3}{2}\cdot\dfrac{4}{13}=\dfrac{-6}{13}\)

c) Ta có: \(C=\dfrac{1}{2}\)

nên \(\dfrac{x}{x^2+1}=\dfrac{1}{2}\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\)(Loại)