Hệ phương trình \(\left\{ \begin{align}  & 3\sqrt{4x+2y}-5\sqrt{2x-y}=2 \\ & 7\sqrt{4x+2y}+2\sqrt{2x-y}=32 \\\end{align} \right.\)có nghiệm là (x; y). Khi đó hiệu \(x-y=?\) 

Cách giải:

\(\left\{ \begin{array}{l}3\sqrt {4x + 2y} - 5\sqrt {2x - y} = 2\\7\sqrt {4x + 2y} + 2\sqrt {2x - y} = 32\end{array} \right.\,\,\,\,\,\,\,\,\,(3)\)

ĐK:  \(4x+2y\ge 0;\,\,2x-y\ge 0\,\,\,\,\,(*).\)

Đặt  \(a=\sqrt{4x+2y}\,,\,b=\sqrt{2x-y}\,\,(a\ge 0,\,b\ge 0)\), khi đó hệ (3) trở thành

\(\begin{array}{l}\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}3a - 5b = 2\\7a + 2b = 32\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}6a - 10b = 4\\35a + 10b = 160\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}6a - 10b = 4\\6a - 10b + 35a + 10b = 4 + 160\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}6a - 10b = 4\\41a = 164\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}10b = 6a - 4\\a = 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 4\\10b = 6.4 - 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 4\\10b = 20\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 4\,(tm)\\b = 2\,(tm)\end{array} \right.\end{array}\)

Thay a = 4, b = 2 ta có :

\(\begin{array}{l}\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}\sqrt {4x + 2y} = 4\\\sqrt {2x - y} = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}4x + 2y = 16\\2x - y = 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}4x + 2(2x - 4) = 16\\y = 2x - 4\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}4x + 4x - 8 = 16\\y = 2x - 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}8x = 24\\y = 2x - 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 3\\y = 2.3 - 4\end{array} \right.\left\{ \begin{array}{l}x = 3\\y = 2\end{array} \right.\end{array}\)

Thay x = 3, y = 2 thì điều kiện (*) được thỏa mãn. Vậy (x; y) = (3; 2) là nghiêm của hệ (3).

Khi đó \(x-y=3-2=1.\)