Question

Giải phương trình : \(2\sqrt{\frac{x^2+x+1}{x+4}}+x^2-4=\frac{2}{\sqrt{x^2+1}}\)

Answer 1

\(2\left(\sqrt{\frac{x^2+x+1}{x+4}}-1\right)+x^2-3=\frac{2}{\sqrt{x^2+1}}-1\)

\(\Leftrightarrow2\frac{\frac{x^2+x+1}{x+4}-1}{\sqrt{\frac{x^2+x+1}{x+4}}+1}+x^2-3=\frac{4-\left(x^2+1\right)}{\left(2+\sqrt{x^2+1}\right)\sqrt{x^2+1}}\)

\(\Leftrightarrow\frac{2\left(x^2-3\right)}{\sqrt{\left(x+4\right)\left(x^2+x+1\right)}+x+4}+x^2-3=\frac{3-x^2}{\left(2\sqrt{x^2+1}\right)\sqrt{x^2+1}}\)

\(\Leftrightarrow\left(x^2-3\right)\left(\frac{2}{\sqrt{\left(x+4\right)\left(x^2+x+1\right)}+x+4}+1+\frac{1}{\left(2+\sqrt{x^2+1}\right)\sqrt{x^2+1}}\right)=0\)

................................................................

(Cũng không chắc _-_ )

Answer 2

bạn làm đúng rồi đấy, mình đăng cho vuii thôi :)))