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\(\sqrt{x+6-4\sqrt{x+2}}-\sqrt{9-4\sqrt{5}}=0\left(đk:x\ge-2\right)\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+2}-2\right)^2}=\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(\Leftrightarrow\left|\sqrt{x+2}-2\right|=\left|\sqrt{5}-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}-2=\sqrt{5}-2\\\sqrt{x+2}-2=2-\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=5\\x+2=21-8\sqrt{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=19-8\sqrt{5}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{3;19-8\sqrt{5}\right\}\)
\(\sqrt{4-\sqrt{4+x}}=x\left(Đkxđ:x\ge-4\right)\)
\(\Leftrightarrow4-\sqrt{4+x}=x^2\)
\(\Leftrightarrow4-x^2=\sqrt{4+x}\)
\(\Leftrightarrow\left(4-x^2\right)^2=4+x\left(đkxđ:x^2\le4\right)\)
\(\Leftrightarrow16-8x^2+x^4=4+x\left(-2\le x\le2\right)\)
\(\Leftrightarrow x^4-8x^2-x+12=0\)
\(\Leftrightarrow\left(x^2-x-4\right)\left(x^2+x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-4=0\\x^2+x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1\pm\sqrt{17}}{2}\\x=\frac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)
Từ: \(Đkxđ:-2\le x\le2\) ta có:
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1\sqrt{17}}{2}\\x=\frac{-1+\sqrt{13}}{2}\end{matrix}\right.\)
Vậy ............
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-6\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-6\right)}=\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-6\right)}\)
\(\Leftrightarrow\dfrac{x-8\sqrt{x}+12}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-6\right)}-\dfrac{x-9\sqrt{x}+20}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-6\right)}=0\)
\(\Leftrightarrow\dfrac{x-8\sqrt{x}+12-x+9\sqrt{x}-20}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-6\right)}=0\)
\(\Leftrightarrow\sqrt{x}-8=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{8}\\x=-\sqrt{8}\end{matrix}\right.\)
C1:\(\sqrt{x+\sqrt{x-4}}+\sqrt{x-\sqrt{x-4}}=0\)
\(\Rightarrow\sqrt{x-4+\sqrt{x-4}+4}+\sqrt{x-4-\sqrt{x-4}+4}=0\)
\(\Rightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=0\)
\(\Rightarrow\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x-4}+2+\sqrt{x-4}-2=0\\\sqrt{x-4}+2+2-\sqrt{x-4}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2\sqrt{x-4}=0\Rightarrow\sqrt{x-4}=0\Rightarrow x-4=0\Rightarrow x=4\\4=0\Rightarrow vôlí\end{matrix}\right.\)
\(\Rightarrow x=4\)
Ta có :
\(\sqrt{4-\sqrt{4+x}}=x\)
\(\Leftrightarrow4-\sqrt{4+x}=x^2\)
\(\Leftrightarrow-\sqrt{4+x}=-4+x^2\)
\(\Leftrightarrow\)\(\sqrt{4+x}=4-x^2\)
\(\Leftrightarrow4+x=16-8x^2+x^4\)
\(\Leftrightarrow4+x-16+8x^2-x^4=0\)
\(\Leftrightarrow-12+x+8x^2-x^4=0\)
\(\Leftrightarrow4x^2+4x-12+x^3+x^2-3x-x^4-x^3+3x^2=0\)
\(\Leftrightarrow4\left(x^2+x-3\right)+x\left(x^2+x-3\right)+x^2\left(x^2+x-3\right)=0\)
\(\Leftrightarrow-\left(x^2-x-4\right)\left(x^2+x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+x-3=0\\x^2-x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1\pm\sqrt{13}}{2}\\x=\frac{1\pm\sqrt{17}}{2}\end{cases}}\)
Kiếm tra lại nghiệm thấy :
\(x=\frac{-1+\sqrt{13}}{2}\)thỏa mãn.
\(x=\frac{-1-\sqrt{13}}{2}\); \(x=\frac{1\pm\sqrt{17}}{2}\)vô lí
Vậy tập nghiệm của phương trình là : \(S=\left\{\frac{-1+\sqrt{13}}{2}\right\}\)