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a,
\(pt\Leftrightarrow\left(x-1-2\sqrt{x-1}+1\right)+\left(y-2-4\sqrt{y-2}+4\right)+\left(z-3-6\sqrt{z-3}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}\)
Ta có:\(x+y+z+35=4\sqrt{x+y}+6\sqrt{y+2}+8\sqrt{z+3}\)
AD BĐT Cô si :
\(\left(x+1\right)+4\ge2\sqrt{\left(x+1\right)4}=2\sqrt{x+1}\)(1)
\(\left(y+2\right)+9\ge2\sqrt{\left(y+2\right)9}=6\sqrt{y+2}\)(2)
\(\left(z+3\right)+16\ge2\sqrt{\left(z+3\right)16}=8\sqrt{z+3}\)(3)
Cộng (1)(2)(3) với nhau ta được:
\(x+y+z+35\ge4\sqrt{x+1}+6\sqrt{y+2}+8\sqrt{z+3}\)
Dấu "=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\y+2=9
\\z+3=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\\z=13\end{matrix}\right.\)
chả cần HĐT dùng Cosi cx đc
\(\left(x+1\right)+4\ge4\sqrt{x+1}\)
\(\left(y+2\right)+9\ge6\sqrt{y+2}\)
\(\left(z+3\right)+16\ge8\sqrt{z+3}\)
\(\Rightarrow VT\ge VP\).Dấu = khi x=3;y=7;z=13
ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ge0\\y+2\ge0\\z+3\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge-1\\y\ge-2\\z\ge-3\end{matrix}\right.\)
Ta có : \(x+y+z+35=2\left(2\sqrt{x+1}+3\sqrt{y+2}+4\sqrt{z+3}\right)\)
=> \(x+y+z+35=4\sqrt{x+1}+6\sqrt{y+2}+8\sqrt{z+3}\)
=> \(x-4\sqrt{x+1}+y-6\sqrt{y+2}+z-8\sqrt{z+3}+35=0\)
=> \(x+1-2.2\sqrt{x+1}+4+y+2-2.3\sqrt{y+2}+9+z+3-4.2\sqrt{z+3}+16=0\)
=> \(\left(\sqrt{x+1}-2\right)^2+\left(\sqrt{y+2}-3\right)^2+\left(\sqrt{z+3}-4\right)^2=0\)
Ta thấy : \(\left\{{}\begin{matrix}\left(\sqrt{x+1}-2\right)^2\ge0\\\left(\sqrt{y+2}-3\right)^2\ge0\\\left(\sqrt{z+3}-4\right)^2\ge0\end{matrix}\right.\)
=> \(\left(\sqrt{x+1}-2\right)^2+\left(\sqrt{y+2}-3\right)^2+\left(\sqrt{z+3}-4\right)^2\ge0\)
- Dấu "=" xảy ra
<=> \(\left\{{}\begin{matrix}\sqrt{x+1}-2=0\\\sqrt{y+2}-3=0\\\sqrt{z+3}-4=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{y+2}=3\\\sqrt{z+3}=4\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x+1=4\\y+2=9\\z+3=16\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=3\\y=7\\z=13\end{matrix}\right.\) ( TM )
Vậy ...
\(ĐKXĐ:\left\{{}\begin{matrix}x\ge-1\\y\ge-2\\z\ge-3\end{matrix}\right.\)
\(PT\Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(\sqrt{y+2}-3\right)^2+\left(\sqrt{z+3}-4\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{y+2}=3\\\sqrt{z+3}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\\z=13\end{matrix}\right.\)
2/ \(3\sqrt[3]{\left(x+y\right)^4\left(y+z\right)^4\left(z+x\right)^4}=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\sqrt[3]{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(\ge6\left(x+y\right)\left(y+z\right)\left(z+x\right)\sqrt[3]{xyz}\)
\(\ge6.\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\sqrt[3]{xyz}\)
\(\ge\frac{16}{3}\left(x+y+z\right)3\sqrt[3]{x^2y^2z^2}\sqrt[3]{xyz}=16xyz\left(x+y+z\right)\)
3/ \(\hept{\begin{cases}\sqrt{xy}+\sqrt{1-x}\le\sqrt{x}\\2\sqrt{xy-x}+\sqrt{x}=1\end{cases}}\)
Dễ thấy
\(\hept{\begin{cases}0\le x\le1\\y\ge1\end{cases}}\)
Từ phương trình đầu ta có:
\(\sqrt{x}-\sqrt{xy}\ge\sqrt{1-x}\ge0\)
\(\Leftrightarrow y\le1\)
Vậy \(x=y=1\)
x=3; y=7; z=13. mạng lác lắm mk ko giải chi tiết đc đâu, thông cảm nha