\(\dfrac{x^3-\left(x-1\right)^3}{\left(4x+3\right)\left(x-5\right)}=\dfrac{7x-1}{4x+3}-\dfrac{x}{x-5}\)
\(\Leftrightarrow\dfrac{x^3-x^3+3x^2-3x+1}{\left(4x+3\right)\left(x-5\right)}=\dfrac{\left(7x-1\right)\left(x-5\right)-x\left(4x+3\right)}{\left(4x+3\right)\left(x-5\right)}\)
\(\Leftrightarrow3x^2-3x+1=7x^2-35x-x+5-4x^2-3x\)
\(\Leftrightarrow3x^2-3x+1=3x^2-39x+5\)
\(\Leftrightarrow-3x+1=-39x+5\)
\(\Leftrightarrow36x=4\)
\(\Leftrightarrow x=\dfrac{1}{9}\)
Vậy...
\(\dfrac{x^3-\left(x-1\right)^3}{\left(4x+3\right)\left(x-5\right)}=\dfrac{7x-1}{4x+3}-\dfrac{x}{x-5}\) ( đk: x ≠ \(\dfrac{-3}{4}\) ; x ≠ 5 )
\(\Leftrightarrow\) \(x^3-\left(x-1\right)^3=\left(7x-1\right)\left(x-5\right)-x\left(4x+3\right)\)
\(\Leftrightarrow\) \(\left(x-x+1\right)\left(x^2+x^2-x+x^2-2x+1\right)=7x^2-35x-x+5-4x^2-3x\)\(\Leftrightarrow\) \(3x^2-3x+1=3x^2-39x+5\)
\(\Leftrightarrow\) \(36x=4\)
\(\Leftrightarrow\) \(x=\dfrac{1}{9}\)( TM )
\(S=\left\{\dfrac{1}{9}\right\}\)
