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

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

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

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

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

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

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

Giúp em với ạ! Em cảm ơn nhiềuuu

ĐKXĐ: ....

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

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

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

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

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

\(\Leftrightarrow\left(x+1-2.2\sqrt{x+1}+4\right)+\left(y+2-2.3\sqrt{y+2}+9\right)+\left(z+3-2.4\sqrt{z+3}+16\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x+1}-2\right)^2=0\\\left(\sqrt{y+2}-3^{ }\right)^2=0\\\left(\sqrt{z+3}-4\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\\z=13\end{matrix}\right.\)

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

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

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

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

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