bài 1 : ta có : \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{1}=1\)
bài 2 : ta có : \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\dfrac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow x+y+z=2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}\)
\(\Leftrightarrow x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-2\sqrt{y-1}\right)+\left(z-1-2\sqrt{z-2}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\) vậy ...
