ĐKXĐ : \(x\ge0\)và \(y\ge1\)

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

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

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

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

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\left(\sqrt{y-1}-3\right)^2=0\end{cases}}\)( Vì \(\left(\sqrt{x}-2\right)^2\ge0\forall x\) và \(\left(\sqrt{y-1}-3\right)^2\ge0\forall y\))

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}-2=0\\\sqrt{y-1}-3=0\end{cases}}\)

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

\(\Leftrightarrow\hept{\begin{cases}x=4\\y-1=9\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=4\\y=10\end{cases}}\)

Vậy ...

Ta có : 

\(\sqrt{2}=1,41....\)

\(\sqrt{3}=1,73....\)

\(\Rightarrow\sqrt{2}< \sqrt{3}\)

\(\sqrt{3}\)và \(\sqrt{2}\)

3>2

\(\left(5+\sqrt{3}\right)\left(1-\sqrt{3}\right)-\left(1-2\sqrt{3}\right)^2\)

\(=2-4\sqrt{3}-\left(1-4\sqrt{3}+12\right)\)

\(=2-4\sqrt{3}-1+4\sqrt{3}-12\)

\(=-11\)

\(\left(5+\sqrt{3}\right)\left(1-\sqrt{3}\right)-\left(1-2\sqrt{3}\right)^2\)

=\(5-5\sqrt{3}+\sqrt{3}-3-\left(1-4\sqrt{3}+12\right)\)

=\(5-5\sqrt{3}+\sqrt{3}-3-1+4\sqrt{3}-12\)

=-11

ĐKXĐ: \(x\ge1\)

Bình phương hai vế

\(\Rightarrow x-1+x+1-2\sqrt{\left(x-1\right)\left(x+1\right)}=4\)

\(2x-2\sqrt{x^2-1}=4\)

\(\Leftrightarrow\sqrt{x^2-1}=x-2\)

Bình phương 2 vế 

\(\Rightarrow x^2-1=x^2-4x+4\)

\(\Leftrightarrow4x=5\)

\(x=\frac{5}{4}\left(tm\right)\)

Vậy ...

phân tích thành n. tử

\(a\sqrt{b}-b\sqrt{a}\)

\(=\sqrt{ab}.\left(\sqrt{a}-\sqrt{b}\right)\)

Có sai sót gì xin bỏ qua~!