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Theo BĐT Am-GM ta có:
\(\sqrt{x-a}=\sqrt{\left(x-a\right).1}\le\frac{x-a+1}{2}\)
\(\sqrt{y-b}\le\frac{y-b+1}{2}\)
\(\sqrt{z-c}\le\frac{z-c+1}{2}\)
Do đó \(VT\le\frac{1}{2}\left(x+y+z\right)\)
Dấu = khi \(\hept{\begin{cases}x=a+1\\y=b+1\\z=c+1\end{cases}}\)
Vậy pt có nghiệm \(\left(x;y;z\right)=\left(a+1;b+1;c+1\right)\)
\(\sqrt{x-a}\le\frac{x-a+1}{2}\Rightarrow\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}\le\frac{x+y+z-a-b-c+3}{2}=\frac{1}{2}\left(x+y+z\right)\)
Dấu = xảy ra khi và chỉ khi : x=a+1 , y=b+1 , z=c+1.
ĐKXĐ: ...
\(\Leftrightarrow2\sqrt{x-a}+2\sqrt{y-b}+2\sqrt{z-c}=x+y+z\)
\(\Leftrightarrow x-a-2\sqrt{x-a}+1+y-b-2\sqrt{y-b}+1+z-c-2\sqrt{z-c}+1+\left(a+b+c-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-a}-1\right)^2+\left(\sqrt{y-b}-1\right)^2+\left(\sqrt{z-c}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}=1\\\sqrt{y-b}=1\\\sqrt{z-c}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=a+1\\y=b+1\\z=c+1\end{matrix}\right.\)
\(pt\Leftrightarrow\left(x-a-2\sqrt{x-a}+1\right)+\left(y-b-2\sqrt{y-b}+1\right)+\left(z-c-\sqrt{z-c}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-a}-1\right)^2+\left(\sqrt{y-b}-1\right)^2+\left(\sqrt{z-c}-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-a}-1=0\\\sqrt{y-b}-1=0\\\sqrt{z-c}-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=a+1\\y=b+1\\z=c+1\end{cases}}\)
Điều kiện: x \(\ge\) a ; y \(\ge\)b ; z \(\ge\) c và x + y + z \(\ge\) 0
PT <=> \(2\sqrt{x-a}+2\sqrt{y-b}+2\sqrt{z-c}=x +y+z\)
<=> \(x-2\sqrt{x-a}+y-2\sqrt{y-b}+z-2\sqrt{z-c}=0\)
<=> \(\left(x-a-2\sqrt{x-a}+1\right)+\left(y-b-2\sqrt{y-b}+1\right)+\left(z-c-2\sqrt{z-c}+1\right)=0\) (Vì a+ b +c = 3)
<=> \(\left(\sqrt{x-a}-1\right)^2+\left(\sqrt{y-b}-1\right)^2+\left(\sqrt{z-c}-1\right)^2=0\)
<=> \(\left(\sqrt{x-a}-1\right)^2=\left(\sqrt{y-b}-1\right)^2=\left(\sqrt{z-c}-1\right)^2=0\)
<=> \(\sqrt{x-a}-1=\sqrt{y-b}-1=\sqrt{z-c}-1=0\)
<=> x - a = 1 ; y - b = 1 ; z - c = 1
<=> x = a+ 1; y = b + 1; z = c+ 1 (Thỏa mãn ĐK)
Vậy....