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Ta có : \(x+y+z+5=2\sqrt{x-1}+4\sqrt{y-3}+6\sqrt{z-5}\left(DK:x\ge1;y\ge3;z\ge5\right)\)
\(\Leftrightarrow\left[\left(x-1\right)-2\sqrt{x-1}+1\right]+\left[\left(y-3\right)-4\sqrt{y-3}+4\right]+\left[\left(z-5\right)-6\sqrt{z-5}+9\right]=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{x-1}-1\right)^2=0\\\left(\sqrt{y-3}-2\right)^2=0\\\left(\sqrt{z-5}-3\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=7\\z=14\end{cases}}\)(TMDK)
Vậy nghiệm của phương trình là : \(\left(x;y;z\right)=\left(2;7;14\right)\)
a,
\(pt\Leftrightarrow\left(x-1-2\sqrt{x-1}+1\right)+\left(y-2-4\sqrt{y-2}+4\right)+\left(z-3-6\sqrt{z-3}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}\)
\(\sqrt{x-y+z}=\sqrt{x}-\sqrt{y}+\sqrt{z}\)
Điều kiện tự làm nhé
\(\Leftrightarrow x-y+z=x+y+z+2\left(\sqrt{xz}-\sqrt{xy}-\sqrt{yz}\right)\)
\(\Leftrightarrow y+\sqrt{xz}-\sqrt{xy}-\sqrt{yz}\)
\(\Leftrightarrow\left(\sqrt{z}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=y\\z=y\end{cases}}\)
\(\Leftrightarrow x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)
\(\Leftrightarrow\left[\left(\sqrt{x}\right)^2-2.\sqrt{x}.1+1^2\right]+\left[\left(\sqrt{y-1}\right)^2+2.\sqrt{y-1}.1+1^2\right]+\left[\left(\sqrt{z-2}\right)^2+2.\sqrt{z-x}.1+1^2\right]-1+1=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-1=0\)
\(\sqrt{y-1}-1=0\)
\(\sqrt{z-2}-1=0\)
\(\Leftrightarrow x=1;y=2;z=3\)
