a) Giải phương trình ( căn bậc hai x+ 23 - căn bậc hai x+ 7 ) ( căn bậc hai 6-x + 2)=8

a)ĐK: -7\(\; \le \;\)x\(\; \le \) 6

Với đk trên thì \(\sqrt {x + 23}  + \sqrt {x + 7} \) \( \ne \) 0

Do đó (*)\(\; \Leftrightarrow \) 16\(\left( {\sqrt {6 - x}  + 2} \right)\) = 8\(\left( {\sqrt {x + 23}  - \;\sqrt {x + 7} } \right)\)

                \( \Leftrightarrow \) 2\(\left( {\sqrt {6 - x}  + 2} \right)\) = \(\sqrt {x + 23}  - \;\sqrt {x + 7} \)

  \( \Leftrightarrow \) \(\sqrt {x + 23} \) – 5 +\(\;\sqrt {x + 7} \) – 3 + 2\(\left( {2 - \sqrt {6 - x} } \right)\) = 0

  \( \Leftrightarrow \) \(\frac{{x\; - \;2}}{{\sqrt {x + 23} {\rm{\;}} + {\rm{\;}}5{\rm{\;}}}}\) + \(\frac{{x\; - \;2}}{{\sqrt {x + 7} {\rm{\;}} + {\rm{\;}}3{\rm{\;}}}}\) + 2.\(\;\frac{{x\; - \;2}}{{\left( {2\; + \;\sqrt {6 - x} } \right){\rm{\;}}}}\) = 0

\(\;\; \Leftrightarrow \) (x – 2)\(\left( {\frac{1}{{\sqrt {x + 23} {\rm{\;}} + {\rm{\;}}5{\rm{\;}}}} + \frac{1}{{\sqrt {x + 7} {\rm{\;}} + {\rm{\;}}3{\rm{\;}}}} + \frac{2}{{\left( {2\; + \;\sqrt {6 - x} } \right){\rm{\;}}}}} \right)\) = 0

\(\;\; \Leftrightarrow \) x – 2 = 0 ( do \(\frac{1}{{\sqrt {x + 23} {\rm{\;}} + {\rm{\;}}5{\rm{\;}}}} + \frac{1}{{\sqrt {x + 7} {\rm{\;}} + {\rm{\;}}3{\rm{\;}}}} + \frac{2}{{\left( {2\; + \;\sqrt {6 - x} } \right){\rm{\;}}}}\) \( > \) 0)

\(\;\; \Leftrightarrow \) x = 2 (t/m đk)

Vậy PT đã cho có nghiệm duy nhất x = 2

b)ĐK: x\(\; \ne \;\)0 ; y\(\; \ne \;\)0

Đặt a = \(x + \frac{1}{y}\) , b = \(y + \frac{1}{x}\)

HPT đã cho trở thành \(\left\{ {\begin{array}{*{20}{c}}{a + b = \frac{9}{2}\;\;\;\;\;\;\;\;\left( 1 \right)}\\{\frac{9}{4} + \frac{3}{2}a = ab\;\;\;\;\left( 2 \right)}\end{array}} \right.\)

Từ (1): b = \(\frac{9}{2} - a\). Thay vào (2):

\(\frac{9}{4} + \frac{3}{2}a\) = a\(\left( {\frac{9}{2} - a} \right)\) \( \Leftrightarrow \) 9 + 6a = 2a (9 – 2a)

\( \Leftrightarrow \) 4a2 – 12a + 9 = 0 \( \Leftrightarrow \;{\left( {2a - 3} \right)^2}\) = 0

\( \Leftrightarrow \) 2a – 3 = 0 \( \Leftrightarrow \;\)a = \(\frac{3}{2}\) \( \Rightarrow \) b = 3

Vậy  \(\left\{ {\begin{array}{*{20}{c}}{x + \frac{1}{y} = \frac{3}{2}}\\{y + \frac{1}{x} = 3}\end{array}} \right.\) \( \Rightarrow \) \(\left\{ {\begin{array}{*{20}{c}}{2xy + 2 = 3y\;\;\;\;\left( 3 \right)}\\{xy + 1 = 3x\;\;\;\;\;\;\left( 4 \right)}\end{array}} \right.\)

\( \Rightarrow \) y = 2x. Thay vào (4):

2x2 – 3x + 1 = 0 \( \Leftrightarrow \) \(\left[ {\begin{array}{*{20}{c}}{x = 1}\\{x = \frac{1}{2}}\end{array}} \right.\)

x = 1 \( \to \) y = 2

x = \(\frac{1}{2}\) \( \to \) y = 1              (t/m đk)

Vậy (x;y)  \(\left\{ {\left( {1;2} \right);\left( {\frac{1}{2};1} \right)} \right\}\)