Question

\(\dfrac{x^3-\left(x-1\right)^3}{\left(4x+3\right)\left(x-5\right)}\)=\(\dfrac{7x-1}{4x+3}\)\(-\dfrac{x}{x-5}\)

Answer 1

\(\dfrac{x^3-\left(x-1\right)^3}{\left(4x+3\right)\left(x-5\right)}=\dfrac{7x-1}{4x+3}-\dfrac{x}{x-5}\left(x\ne-\dfrac{3}{4};x\ne5\right)\\ \Leftrightarrow\dfrac{x^3-\left(x-1\right)^3}{\left(4x+3\right)\left(x-5\right)}=\dfrac{\left(7x-1\right)\left(x-5\right)}{\left(4x+3\right)\left(x-5\right)}-\dfrac{x\left(4x+3\right)}{\left(4x+3\right)\left(x-5\right)}\\ \Leftrightarrow x^3-\left(x-1\right)^3=\left(7x-1\right)\left(x-5\right)-x\left(4x+3\right)\\ \Leftrightarrow x^3-x^3+3x^2-3x+1=7x^2-35x-x+5-4x^2-3x\\ \Leftrightarrow3x^2-3x+1=3x^2-39x+5\\ \Leftrightarrow-3x+1=-39x+5\\ \Leftrightarrow-3x+39x=5-1\\ \Leftrightarrow36x=4\\ \Leftrightarrow x=\dfrac{4}{36}=\dfrac{1}{9}\left(tm\right)\)