Question

Tìm m để \(\left(x+1\right)^2+\sqrt{2x\left(x+a+1\right)}=a^2+1+\left|x+a\right|\) có đúng 1 nghiệm thuộc [-2;2]

Answer 1

\(\left(x+1\right)^2+\sqrt{2x\left(x+a+1\right)}=a^2+1+\left|x+a\right|\)

\(\Leftrightarrow x^2+2x+\sqrt{2x^2+2xa+2x}=a^2+\left|x+a\right|\)

\(\Leftrightarrow2x^2+2ax+2x+\sqrt{2x^2+2xa+2x}=x^2+2ax+a^2+\left|x+a\right|\)

\(\Leftrightarrow\left(\sqrt{2x^2+2xa+2x}+\dfrac{1}{2}\right)^2=\left(\left|x+a\right|+\dfrac{1}{2}\right)^2\)

\(\Leftrightarrow\sqrt{2x^2+2xa+2x}=\left|x+a\right|\)

\(\Leftrightarrow2x^2+2xa+2x=x^2+2xa+a^2\)

\(\Leftrightarrow x^2+2x=a^2\)

Đồ thị hàm số:

Dựa vào đồ thị, yêu cầu bài toán thỏa mãn khi \(0\le a^2\le8\Leftrightarrow-2\sqrt{2}\le a\le2\sqrt{2}\)