Question

[Châu Hữu Phát]

Giải phương trình 

\(\sqrt{130307+140307\sqrt{1+x}}=1+\sqrt{130307-140307\sqrt{1+x}}\)

Answer 1

[Mr Lazy]

Đặt \(130307=a;\text{ }140307=b\)

Pt trở thành \(\sqrt{a+b\sqrt{x+1}}=1+\sqrt{a-b\sqrt{x+1}}\)

\(\Leftrightarrow\sqrt{a+b\sqrt{x+1}}-\sqrt{a-b\sqrt{x+1}}=1\)

\(\Leftrightarrow a+b\sqrt{x+1}+a-b\sqrt{x+1}-2\sqrt{\left(a+b\sqrt{x+1}\right)\left(a-b\sqrt{x+1}\right)}=1\)

\(\Leftrightarrow2a-1=2\sqrt{a^2-b^2\left(x+1\right)}\)

\(\Leftrightarrow\left(2a-1\right)^2=4\left[a^2-b^2\left(x+1\right)\right]\)

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

\(\Leftrightarrow x=\frac{4a^2-\left(2a-1\right)^2}{4b^2}-1\)