8) ĐKXĐ: \(x\notin\left\{2;\dfrac{4}{3}\right\}\)
Ta có: \(\dfrac{5}{x-2}+\dfrac{6}{3-4x}=0\)
\(\Leftrightarrow\dfrac{5\left(3-4x\right)}{\left(x-2\right)\left(3-4x\right)}+\dfrac{6\left(x-2\right)}{\left(3-4x\right)\left(x-2\right)}=0\)
Suy ra: \(15-20x+6x-12=0\)
\(\Leftrightarrow-14x+3=0\)
\(\Leftrightarrow-14x=-3\)
\(\Leftrightarrow x=\dfrac{3}{14}\)
Vậy: \(S=\left\{\dfrac{3}{14}\right\}\)
10) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}=\dfrac{1}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2=1\)
\(\Leftrightarrow x^2+3x+1=0\)
\(\Delta=9-4\cdot1\cdot1=5\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{5}}{2}\left(nhận\right)\\x_2=\dfrac{-3+\sqrt{5}}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{-3-\sqrt{5}}{2};\dfrac{-3+\sqrt{5}}{2}\right\}\)
8, đk: x≠2; x≠\(\dfrac{3}{4}\)
\(\dfrac{5}{x-2}+\dfrac{6}{3-4x}=0\)
\(\Leftrightarrow\dfrac{5}{x-2}=\dfrac{-6}{3-4x}\)
\(\Leftrightarrow5.\left(3-4x\right)=-6.\left(x-2\right)\)
\(\Leftrightarrow15-20x=-6x+12\)
\(\Leftrightarrow-14x=-3\)
\(\Leftrightarrow x=\dfrac{3}{14}\)( thỏa mãn)
