Question

giải phương trình sau

\(\dfrac{x+5}{3}-\dfrac{x-3}{5}=\dfrac{5}{x-3}-\dfrac{3}{x+5}\)

\(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)

\(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)

Answer 1

b: \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)

\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)+4=x^2+2x-3\)

\(\Leftrightarrow3x^2+8x-3-2x^2-3x+5+4=x^2+2x-3\)

\(\Leftrightarrow x^2+5x+6=x^2+2x-3\)

=>3x=-9

hay x=-3(loại)

c: \(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)

\(\Leftrightarrow x^2+2x-15=x^2-1-8=x^2-9\)

=>2x=6

hay x=3(loại)