\(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=-2x\)
Đặt \(\hept{\begin{cases}\sqrt{4+x}=a\ge0\\\sqrt{4-x}=b\ge0\end{cases}}\) thì ta có:
\(\hept{\begin{cases}\left(a-2\right)\left(b+2\right)=b^2-a^2\left(1\right)\\8=a^2+b^2\left(2\right)\end{cases}}\)
Lấy (2) + 2.(1) vế theo vế rút gọn ta được
\(\Leftrightarrow3b^2-a^2+4b-4a-2ab=0\)
\(\Leftrightarrow\left(b-a\right)\left(3b+a+4\right)=0\)
\(\Leftrightarrow a=b\)
\(\Rightarrow\sqrt{4+x}=\sqrt{4-x}\)
\(\Leftrightarrow x=0\)
Ta có : \(\left(\sqrt{x+4}-2\right)\left(\sqrt{x+4}+2\right)=-2x\)
\(\Rightarrow\left(\sqrt{x+4}\right)^2-2^2=-2x\)
\(\Leftrightarrow x+4-4=-2x\)
=> x = -2x
=> x + 2x = 0
=> 3x = 0
=> x = 0
Vậy x = 0.
\(\left(\sqrt{x+4}-2\right)\) (1)
\(\left(\sqrt{4-x}+2\right)\) (2)
Lấy (1) . (2) \(\Leftrightarrow\left(\sqrt{x+4}-2\right).\left(\sqrt{4-x}+2\right)\Leftrightarrow0.4=-2x\)
Từ (1) và (2) , Ta có: \(0.4=0\Rightarrow x=0\)
