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chuyển vế rồi thêm bớt cậu sẽ có rồi tìm được x=1 y=1 z=4
\(\left(x-2\sqrt{x}+1\right)+\left(y-2\sqrt{y}+1\right)+\left(z-4\sqrt{z}+4\right)=0\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2+\left(\sqrt{z}-2\right)^2=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: \(x\ge-1;y\ge3;z\ge1\)
\(\sqrt{x+1}+\sqrt{y-3}+\sqrt{z-1}=\dfrac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow x+1-2\sqrt{x+1}+1+y-3-2\sqrt{y-3}+1+z-1-2\sqrt{z-1}+1=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2=0\)
Ta thấy: \(\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2\ge0\)
Đẳng thức xảy ra khi:
\(\left\{{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{y-3}=1\\\sqrt{z-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\\z=2\end{matrix}\right.\)
Cách khác:
ĐK: \(x\ge-1;y\ge3;z\ge1\)
Áp dụng BĐT \(ab\le\dfrac{a^2+b^2}{2}\).
\(\sqrt{x+1}\le\dfrac{x+1+1}{2}=\dfrac{x+2}{2}\)
\(\sqrt{y-3}\le\dfrac{y-3+1}{2}=\dfrac{y-2}{2}\)
\(\sqrt{z-1}\le\dfrac{z-1+1}{2}=\dfrac{z}{2}\)
Cộng vế theo vế các BĐT trên ta được:
\(\sqrt{x+1}+\sqrt{y-3}+\sqrt{z-1}\le\dfrac{1}{2}\left(x+y+z\right)\)
Đẳng thức xảy ra khi:
\(\left\{{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{y-3}=1\\\sqrt{z-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\\z=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge-1;y\ge3;z\ge1\)
\(\Leftrightarrow x+y+z-2\sqrt{x+1}-2\sqrt{y-3}-2\sqrt{z-1}=0\)
\(\Leftrightarrow\left(x+1-2\sqrt{x+1}+1\right)+\left(y-3-2\sqrt{y-3}+1\right)+\left(z-1-2\sqrt{z-1}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}-1=0\\\sqrt{y-3}-1=0\\\sqrt{z-1}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\\z=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+y+z+35=2\left(2\sqrt{x+1}+3\sqrt{y+2}+4\sqrt{z+3}\right)\)
\(\Leftrightarrow x+y+z+35-4\sqrt{x+1}-6\sqrt{y+2}-8\sqrt{z+3}=0\)
\(\Leftrightarrow\left(x+1-4\sqrt{x+1}+4\right)+\left(y+2-6\sqrt{y+2}+9\right)+\left(z+3-8\sqrt{z+3}+16\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(\sqrt{y+2}-3\right)^2+\left(\sqrt{z+3}-4\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{x+1}-2\right)^2=0\\\left(\sqrt{y+2}-3\right)^2=0\\\left(\sqrt{z+3}-4\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x+1}=2\\\sqrt{y+2}=3\\\sqrt{z+3}=4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=3\\y=7\\z=13\end{cases}}\)
cộng 3 vế vào với nhau 2(x+y+z)=13/12
x+y+z=13/24
hiệu giữa y và z là 1/4,x với z là 1/6
3z+1/4+1/6=13/24
3z=13/24-5/12
z=1/8 :3=1/24
tự tình nốt nhé Giang yêu quý!