Đề thi chính thức vào 10 môn Toán Sở GD&ĐT Bình Đị...

A \(\begin{array}{l}a)\,A = \frac{{1 - x}}{x}\\b)\,0\, < \,x\, < \frac{2}{3}\end{array}\)

B \(\begin{array}{l}a)\,A = \frac{{1 - x}}{x}\\b)\,0\, < \,x\, < \frac{3}{2}\end{array}\)

C \(\begin{array}{l}a)\,A = \frac{{1 + x}}{x}\\b)\,0\, < \,x\, < \frac{2}{3}\end{array}\)

D \(\begin{array}{l}a)\,A = \frac{{1 + x}}{x}\\b)\,0\, < \,x\, < \,\frac{3}{2}\end{array}\)

A \(\begin{array}{l}1)\,\left( {x;\,y} \right) = \left( {1;\, - 2} \right)\\2)\,a)\,A\left( {\frac{{k + 3}}{k};\,0} \right);\,\,B\left( {0;\, - k - 3} \right)\\b)\,{S_{OAB}} = \frac{{25}}{4}\end{array}\)

B \(\begin{array}{l}1)\,\left( {x;\,y} \right) = \left( {1;\,2} \right)\\2)\,a)\,A\left( {\frac{{k + 3}}{k};\,0} \right);\,\,B\left( {0;\,k - 3} \right)\\b)\,{S_{OAB}} = \frac{{25}}{4}\end{array}\)

C \(\begin{array}{l}1)\,\left( {x;\,y} \right) = \left( { - 1;\,2} \right)\\2)\,a)\,A\left( {\frac{{k + 3}}{k};\,0} \right);\,\,B\left( {0;\, - k - 3} \right)\\b)\,{S_{OAB}} = \frac{{25}}{2}\end{array}\)

D \(\begin{array}{l}1)\,\left( {x;\,y} \right) = \left( {1;\, - 2} \right)\\2)\,a)\,A\left( {\frac{{k + 3}}{k};\,0} \right);\,\,B\left( {0;\, - k + 3} \right)\\b)\,{S_{OAB}} = \frac{{25}}{2}\end{array}\)