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\(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=-2x\)
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.
a) Ta có: \(\left(x-\sqrt{2}\right)+3\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)+3\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(1+3x+3\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{2}=0\\3x+3\sqrt{2}+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\3x=-3\sqrt{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=\dfrac{-3\sqrt{2}-1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{2};\dfrac{-3\sqrt{2}-1}{3}\right\}\)
b) Ta có: \(x^2-5=\left(2x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
\(\Leftrightarrow\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)-\left(2x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=0\)
\(\Leftrightarrow\left(x+\sqrt{5}\right)\left(x-\sqrt{5}-2x+\sqrt{5}\right)=0\)
\(\Leftrightarrow-x\left(x+\sqrt{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\x+\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\sqrt{5}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-\sqrt{5}\right\}\)
Câu a : \(3\sqrt{x-2}-\sqrt{x^2-4}=0\) ( ĐK : \(x\ge2\) )
\(\Leftrightarrow3\sqrt{x-2}-\sqrt{\left(x+2\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(3-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\3-\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=7\left(TM\right)\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=7\)