Tìm x, biết 2/(x-1).(x-3)+5/(x-3).(x-8)+12/(x-8).(x-20)-1/(x-20)=-3/4

\(ĐKXĐ:x\ne1,x\ne3,x\ne8,x\ne20\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow3x=3+4\)

\(\Leftrightarrow3x=7\)

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

Vậy \(x=\dfrac{7}{3}\)