Đặt\(A=\sqrt{x+\sqrt{x^2-4}}-4.\sqrt{x-\sqrt{x^2-4}}\)
\(A^2=x+\sqrt{x^2-4}+16.\left(x-\sqrt{x^2-4}\right)-2.4.\sqrt{x^2-\left(\sqrt{x^2-4}\right)^2}\)
\(A^2=x+\sqrt{x^2-4}+16x-16.\sqrt{x^2-4}-8.\sqrt{x^2-x^2+4}\)
\(A^2=17x-15.\sqrt{x^2-4}-16\)
mình làm đến đây đc thôi, sorry
Dễ thây \(x\ge2\)
\(A=\sqrt{x+\sqrt{x^2-4}}-4\sqrt{x-\sqrt{x^2-4}}\)
\(=\sqrt{\frac{2x+2\sqrt{\left(x+2\right)\left(x-2\right)}}{2}}-4\sqrt{\frac{2x-2\sqrt{\left(x+2\right)\left(x-2\right)}}{2}}\)
\(=\sqrt{\frac{\left(x+2\right)+2\sqrt{\left(x+2\right)\left(x-2\right)}+\left(x-2\right)}{2}}-4\sqrt{\frac{\left(x+2\right)-2\sqrt{\left(x+2\right)\left(x-2\right)}+\left(x-2\right)}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{\left(x+2\right)}+\sqrt{\left(x-2\right)}\right)^2}{2}}-4\sqrt{\frac{\left(\sqrt{\left(x+2\right)}-\sqrt{\left(x-2\right)}\right)^2}{2}}\)
\(=\frac{1}{\sqrt{2}}\left[\left(\sqrt{x+2}+\sqrt{x-2}\right)-4\left(\sqrt{x+2}-\sqrt{x-2}\right)\right]\)
\(=\frac{1}{\sqrt{2}}\left(-3\sqrt{x+2}+5\sqrt{x-2}\right)\)
