Question

\(\dfrac{6}{x^2-1}+5=\dfrac{8x-1}{4x-4}-\dfrac{12x-1}{4-4x}\)

Giaỉ phương trình sau:

Answer 1

học ngu, dễ vậy

Answer 2

Quy đồng, khử mẫu

Answer 3

\(\dfrac{6}{x^2-1}+5=\dfrac{8x-1}{4x-4}-\dfrac{12x-1}{4-4x}\) Đkxđ : x≠1,-1

\(\Leftrightarrow\dfrac{6}{x^2-1}+5=\dfrac{8x-1}{4x-4}+\dfrac{12x-1}{4x-4}\)

\(\Leftrightarrow\dfrac{6}{x^2-1}+5=\dfrac{20x-2}{4x-4}\)

\(\Leftrightarrow\dfrac{5x^2+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{10x-1}{2\left(x-1\right)}\)

\(\Leftrightarrow\dfrac{5x^2+1}{\left(x+1\right)\left(x-1\right)}-\dfrac{10x-1}{2\left(x-1\right)}=0\)

\(\Leftrightarrow\dfrac{1-9x}{2\left(x+1\right)\left(x-1\right)}=0\)

\(\Leftrightarrow1-9x=0\)

\(\Leftrightarrow9x=1\)

\(\Leftrightarrow x=\dfrac{1}{9}\)

Vậy S=\(\left\{\dfrac{1}{9}\right\}\)