Question

Giải phương trình: \(\frac{9}{4(x+4)^2}+1=\frac{8}{(2x+5)^2}\)

Answer 1

ĐKXĐ: ...

\(\Leftrightarrow\frac{9\left(2x+5\right)^2}{4\left(x+4\right)^2}+\left(2x+5\right)^2=8\)

\(\Leftrightarrow\frac{9\left(2x+5\right)^2}{4\left(x+4\right)^2}-2.\frac{3\left(2x+5\right)}{2\left(x+4\right)}.\left(2x+5\right)+\left(2x+5\right)^2+\frac{3\left(2x+5\right)^2}{x+4}=8\)

\(\Leftrightarrow\left(\left(2x+5\right)-\frac{3\left(2x+5\right)}{2\left(x+4\right)}\right)^2+\frac{3\left(2x+5\right)^2}{x+4}=8\)

\(\Leftrightarrow\left(\frac{\left(2x+5\right)^2}{2\left(x+4\right)}\right)^2+\frac{3\left(2x+5\right)^2}{x+4}-8=0\)

Đặt \(\frac{\left(2x+5\right)^2}{x+4}=a\)

\(\Leftrightarrow\frac{a^2}{4}+3a-8=0\)

Nghiệm xấu, bạn tự giải nốt