Đề thi chính thức vào 10 môn Toán Trường Phổ Thông...

A \(P = \left\{ \begin{array}{l}\frac{{\sqrt {{a^2} + {b^2}} }}{{\sqrt {{a^2} - {b^2}} }}\,\,\,khi\,\,\,a > 0\\ - \frac{{\sqrt {{a^2} + {b^2}} }}{{\sqrt {{a^2} - {b^2}} }}\,\,\,khi\,\,\,a < 0\end{array} \right.\)

B \(P = \left\{ \begin{array}{l}\frac{{\sqrt {{a^2} - {b^2}} }}{{\sqrt {{a^2} + {b^2}} }}\,\,\,khi\,\,\,a > 0\\ - \frac{{\sqrt {{a^2} - {b^2}} }}{{\sqrt {{a^2} + {b^2}} }}\,\,\,khi\,\,\,a < 0\end{array} \right.\)

C \(P = \left\{ \begin{array}{l}\frac{{a\sqrt {{a^2} + {b^2}} }}{{b\sqrt {{a^2} - {b^2}} }}\,\,\,khi\,\,\,a > 0\\ - \frac{{a\sqrt {{a^2} + {b^2}} }}{{b\sqrt {{a^2} - {b^2}} }}\,\,\,khi\,\,\,a < 0\end{array} \right.\)

D \(P = \left\{ \begin{array}{l}\frac{{\sqrt {{a^2} - {b^2}} }}{{b\sqrt {{a^2} + {b^2}} }}\,\,\,khi\,\,\,a > 0\\ - \frac{{\sqrt {{a^2} - {b^2}} }}{{b\sqrt {{a^2} + {b^2}} }}\,\,\,khi\,\,\,a < 0\end{array} \right.\)

A \(\left( {a;\;b} \right) = \left\{ {\left( {1;6} \right),\;\left( { - 1; - 6} \right)} \right\}.\)

B \(\left( {a;\;b} \right) = \left\{ {\left( {1;6} \right),\;\left( { - 1;6} \right)} \right\}.\)

C \(\left( {a;\;b} \right) = \left\{ {\left( {1; - 6} \right),\;\left( { - 1;6} \right)} \right\}.\)

D \(\left( {a;\;b} \right) = \left\{ {\left( {1; - 6} \right),\;\left( { - 1; - 6} \right)} \right\}.\)

A \(x = 1\)

B \(x = 2\)

C \(x =  - 1\)

D \(x = 0\)

A \(\min P = 0\,\,;\,\,\,\max P = \frac{5}{2}\)

B \(\min P = 0\,\,;\,\,\,\max P = \frac{7}{2}\)

C \(\min P = 1\,\,;\,\,\,\max P = \frac{5}{2}\)

D \(\min P = 1\,\,;\,\,\,\max P = \frac{7}{2}\)

A \(\left( {x;\;y} \right) = \left( {3;\;2} \right).\)

B \(\left( {x;\;y} \right) = \left( {2;\;3} \right).\)

C \(\left( {x;\;y} \right) = \left( {1;\;2} \right).\)

D \(\left( {x;\;y} \right) = \left( {2;\;1} \right).\)