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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 ...

\(ĐKXĐ:\left\{{}\begin{matrix}x\ge-1\\y\ge-2\\z\ge-3\end{matrix}\right.\)

\(PT\Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(\sqrt{y+2}-3\right)^2+\left(\sqrt{z+3}-4\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{y+2}=3\\\sqrt{z+3}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\\z=13\end{matrix}\right.\)

\(x+y+z+35=2\left(2\sqrt{x+1}+3\sqrt{y+2}+4\sqrt{z+3}\right)\)

\(\Leftrightarrow x+y+z+35-4\sqrt{x+1}-6\sqrt{y+2}-8\sqrt{z+3}=0\)

\(\Leftrightarrow\left(x+1-4\sqrt{x+1}+4\right)+\left(y+2-6\sqrt{y+2}+9\right)+\left(z+3-8\sqrt{z+3}+16\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(\sqrt{y+2}-3\right)^2+\left(\sqrt{z+3}-4\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{x+1}-2\right)^2=0\\\left(\sqrt{y+2}-3\right)^2=0\\\left(\sqrt{z+3}-4\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x+1}=2\\\sqrt{y+2}=3\\\sqrt{z+3}=4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3\\y=7\\z=13\end{cases}}\)

ĐKXĐ: ....

\(x+1-4\sqrt{x+1}+4+y+2-6\sqrt{y+2}+9+z+3-8\sqrt{z+3}+16=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(\sqrt{y+2}-3\right)^2+\left(\sqrt{z+3}-4\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{y+2}=3\\\sqrt{z+3}=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=7\\z=13\end{matrix}\right.\)

\(\Leftrightarrow x+y+z+35=4\sqrt{x+1}+6\sqrt{y+2}+8\sqrt{z+3}\)

\(\Leftrightarrow x+1-4\sqrt{x+1}+4+y+2-6\sqrt{y+2}+9+z+3-8\sqrt{z+3}+16=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(\sqrt{y+2}-3\right)^2+\left(\sqrt{z+3}-4\right)^2=0\)

cho từng cái ngoặc bằng 0 thì ta được x=3 ; y=7 ;z=13         nếu đúng tick nha bạn