Question

[Le DuyDucc_]

Giúp mình với

Answer 1

[Yuri Sweet[𝕿𝖊𝖆𝖒 𝕹𝖊𝖕𝖆𝖑]]

a) ĐKXĐ :\(\hept{\begin{cases}x\ge0\\y\ge1\end{cases}}\)

Ta có : \(x+y+12=4\sqrt{x}+6\sqrt{y-1}\)

\(\Leftrightarrow x+y+12-4\sqrt{x}-6\sqrt{y-1}=0\)

\(\Leftrightarrow\left(x-4\sqrt{x}+4\right)+\left[\left(y-1\right)-6\sqrt{y-1}+9\right]=0\)

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

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}-2=0\\\sqrt{y-1}-3=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\\\sqrt{y-1}=3\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=4\\y=8\end{cases}\left(TM\right)}\)

Vậy PT có 1 nghiệm duy nhất là \(\left(x;y\right)=\left(4;8\right)\)

Answer 2

[Đặng Ngọc Quỳnh]

a) đk: \(x\ge0;y\ge1\)

pt \(\Leftrightarrow x+y+12-4\sqrt{x}-6\sqrt{y-1}=0\)

\(\Leftrightarrow\left(x-4\sqrt{x}+4\right)+\left[\left(y-1\right)-6\sqrt{y-1}+9\right]=0\)

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

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\left(\sqrt{y-1}-3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x}-2=0\\\sqrt{y-1}-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=4\\y=10\end{cases}}}\)

b) đk: \(x\ge-1;y\ge-2;z\ge-3\)

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

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

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