Question

Giải phương trình:

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

Answer 1

\(ĐKXĐ:x\ne\frac{3}{2}\)

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

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

\(\Leftrightarrow x^2+2x+7=x^2+10\)

\(\Leftrightarrow2x=3\)

\(\Leftrightarrow x=\frac{3}{2}\left(KTMĐKXĐ\right)\)

Vậy phương trình vô nghiệm

Answer 2

ĐKXĐ: x khác 3/2

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

<=> \(\frac{x^2+4x+4}{2x-3}-1=\frac{x^2+10}{2x-3}\)

<=> x^2 + 4x + 4 - 2x + 3 = x^2 + 10

<=> x^2 + 4x + 4 - 2x + 3 - x^2 - 10 = 0

<=> 2x - 3 = 0

<=> 2x = 0 + 3

<=> 2x = 3

<=> x = 3 (ktmdk)

=> pt no