Question

Giải pt sau\(x^2-2x+8-4\sqrt{\left(4-x\right)\left(x+2\right)}\)=0

Answer 1

Lời giải:
ĐK: \(-2\leq x\leq 4\)

Ta có: \(x^2-2x+8-4\sqrt{(4-x)(x+2)}=0\)

\(\Leftrightarrow x^2-2x+8-4\sqrt{2x+8-x^2}=0\)

\(\Leftrightarrow 16-(2x-x^2+8)-4\sqrt{2x+8-x^2}=0\)

Đặt \(\sqrt{2x+8-x^2}=t\)

\(\Rightarrow 16-t^2-4t=0\)

\(\Rightarrow t=-2\pm 2\sqrt{5}\). Vì \(t\geq 0\Rightarrow t=-2+2\sqrt{5}\)

\(\Rightarrow t^2=2x+8-x^2=24-8\sqrt{5}\)

\(\Leftrightarrow x^2-2x+16-8\sqrt{5}=0\)

\(\Rightarrow x=1\pm \sqrt{8\sqrt{5}-15}\) (đều thỏa mãn)

Vậy............