Question

Giải phương trình:

\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3}\)

Answer 1

https://i.imgur.com/7FB0wsf.jpg

Answer 2

ĐKXĐ: \(x\ne\frac{3}{2}\)

Ta có: \(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3}\)

\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}-\frac{x^2+10}{2x-3}=0\)

\(\Leftrightarrow x^2+4x+4-\left(2x-3\right)-\left(x^2+10\right)=0\)

\(\Leftrightarrow x^2+4x+4-2x+3-x^2-10=0\)

\(\Leftrightarrow2x-3=0\)

\(\Leftrightarrow2x=3\)

hay \(x=\frac{3}{2}\)(ktm)

Vậy: \(x\in\varnothing\)