Question

help me plsssssssssss

Answer 1

a: ĐKXĐ: \(x\notin\left\{5;8\right\}\)

\(\dfrac{6}{x-5}+\dfrac{2}{x-8}=\dfrac{18}{\left(x-5\right)\left(8-x\right)}-1\)

=>\(\dfrac{6x-48+2x-10}{\left(x-5\right)\left(x-8\right)}=\dfrac{-18}{\left(x-5\right)\left(x-8\right)}-\dfrac{\left(x-5\right)\left(x-8\right)}{\left(x-5\right)\left(x-8\right)}\)

=>\(8x-58=-18-\left(x^2-13x+40\right)\)

=>\(8x-58+18+\left(x^2-13x+40\right)=0\)

=>\(x^2-13x+40+8x-40=0\)

=>\(x^2-5x=0\)

=>x(x-5)=0

=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=5\left(loại\right)\end{matrix}\right.\)

b: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

\(\dfrac{x^2-x}{x+3}-\dfrac{x}{x-3}=\dfrac{7x^2-3x}{9-x^2}\)

=>\(\dfrac{\left(x^2-x\right)\left(x-3\right)-x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{-7x^2+3x}{\left(x-3\right)\left(x+3\right)}\)

=>\(x^3-3x^2-x^2+3x-x^2-3x=-7x^2+3x\)

=>\(x^3-5x^2+7x^2-3x=0\)

=>\(x^3+2x^2-3x=0\)

=>\(x\left(x+3\right)\left(x-1\right)=0\)

=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-3\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)