Giải hệ phương trình: $\left\{ \begin{array}{l} \sqrt{x-1}+\sqrt{2-y}=1  (1)\\ 3\log_9(9x^2)-\log_3 y^3=3  (2) \end{array} \right.$
ĐK: $ \left\{ \begin{array}{l} x\geq 1\\0<y\leq 2 \end{array} \right.$   
$(2)\Leftrightarrow 3(1+\log_3x)-3\log_3y=3\Leftrightarrow \log_3x=\log_3y\Leftrightarrow x=y.$   
Thay $y=x$ vào (1) ta có:   
$\sqrt{x-1}+\sqrt{2-x}=1\Leftrightarrow x-1+2-x+2\sqrt{(x-1)(2-x)}=1$   
$\Leftrightarrow \sqrt{(x-1)(2-x)}=0 \Leftrightarrow\left[ {\begin{matrix}  x=1  \\   x=2 \end{matrix}} \right..$   
Vậy hệ có hai nghiệm là: $(x;y)=(1;1)$ và $(x;y)=(2;2)$.
Điều kiện: $\left\{ \begin{array}{l} x\ge 1\\0< y\le2 \end{array} \right.$.
Hệ phương trình tương đương với:
$\left\{ \begin{array}{l} \sqrt{x-1}+\sqrt{2-y}=1\\3(1+\log_9(x^2))-3\log_3y=3 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l}\sqrt{x-1}+\sqrt{2-y}=1 \\ \log_3x=\log_3y \end{array} \right.$
                                                               $\Leftrightarrow \left\{ \begin{array}{l} x=y\\ \sqrt{x-1}+\sqrt{2-y}=1 \end{array} \right.$
                                                               $\Leftrightarrow \left\{ \begin{array}{l} x=y\\ \sqrt{x-1}+\sqrt{2-x}=1 \end{array} \right.$
                                                               $\Leftrightarrow \left\{ \begin{array}{l} x=y\\ 1+2\sqrt{(x-1)(2-x)}=1 \end{array} \right.$
                                                               $\Leftrightarrow \left\{ \begin{array}{l} x=y\\ (x-1)(2-x)=0 \end{array} \right.$
                                                               $\Leftrightarrow \left [ \begin{array}{l} x=y=1\\ x=y=2 \end{array} \right.$      (thỏa mãn điều kiện)
Vậy nghiệm của hệ là: $(x,y)\in\{(1;1),(2;2)\}$
lời giải hoàn toàn chính xác, trình bày khoa học, chất lượng thấp ở đâu? –  nohssiw 19-07-12 11:16 AM

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