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\(\Leftrightarrow x^2-2x-m+\dfrac{2\left(x^2-2x-m\right)\left(\sqrt{x}+1\right)}{x+\sqrt{2x+m}}=0\)
\(\Leftrightarrow\left(x^2-2x-m\right)\left(1+\dfrac{2\left(\sqrt{x}+1\right)}{x+\sqrt{2x+m}}\right)=0\)
\(\Leftrightarrow x^2-2x-m=0\)
\(\Leftrightarrow\sqrt{\left(x+\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2}-\sqrt{\left(x-\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2}=m\)
Trong mp tọa độ, gọi \(A\left(-\dfrac{1}{2};\dfrac{\sqrt{3}}{2}\right)\) ; \(B\left(\dfrac{1}{2};\dfrac{\sqrt{3}}{2}\right)\) và \(M\left(x;0\right)\) \(\Rightarrow AB=1\)
\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AM}=\left(x+\dfrac{1}{2};-\dfrac{\sqrt{3}}{2}\right)\\\overrightarrow{BM}=\left(x-\dfrac{1}{2};\dfrac{\sqrt{3}}{2}\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}AM=\sqrt{\left(x+\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2}\\BM=\sqrt{\left(x-\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2}\end{matrix}\right.\)
Theo BĐT tam giác: \(\left|AM-BM\right|< AB=1\)
\(\Rightarrow\left|m\right|< 1\Rightarrow-1< m< 1\)
a, ĐK: \(x\le-1,x\ge3\)
\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x^2-2x-3}+3\right).\left(\sqrt{x^2-2x-3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-2x-3}=-\dfrac{3}{2}\left(l\right)\\\sqrt{x^2-2x-3}=1\end{matrix}\right.\)
\(\Leftrightarrow x^2-2x-3=1\)
\(\Leftrightarrow x^2-2x-4=0\)
\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)
b, ĐK: \(-2\le x\le2\)
Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)
Khi đó phương trình tương đương:
\(3t-t^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2+x}-2\sqrt{2-x}=0\\\sqrt{2+x}-2\sqrt{2-x}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)
Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm
Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)