Question

[Đàm Tùng Vận]

Giaỉ phương trình sau

\(\dfrac{1}{2x-3}+\dfrac{1}{2x+3}=\dfrac{2x+4}{4x^2-9}\)

Answer 1

[Tiếng Anh]

\(ĐK:x\ne\pm\dfrac{3}{2}\\ PT\Leftrightarrow2x+3+2x-3=2x+4\\ \Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\)

Answer 2

[Nguyễn Hoàng Tùng]

\(\dfrac{1}{2x-3}+\dfrac{1}{2x+3}=\dfrac{2x+4}{4x^2-9}\)

\(\dfrac{2x+3+2x-3}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{2x+4}{4x^2-9}\)

\(\dfrac{4x}{4x^2-9}=\dfrac{2x+4}{4x^2-9}\Rightarrow4x=2x+4\)

\(\Rightarrow2x=4\Rightarrow x=2\)