Giải phương trình và hệ phương trình sau:
\(a) \; \;2{x^2} - 5x + 2 = 0. \) \(b) \; \left \{ \begin{array}{l}2x + y = 1 \ \3x - 2y = 5 \end{array} \right.. \)
A.\(\begin{array}{l}a)\,S = \left\{ { - \frac{1}{2};\, - 2} \right\}\\b)\,\left( {x;\,y} \right) = \left( {1;\, - 1} \right)\end{array}\)
B.\(\begin{array}{l}a)\,S = \left\{ {\frac{1}{2};\,2} \right\}\\b)\,\left( {x;\,y} \right) = \left( {1;\, - 1} \right)\end{array}\)
C.\(\begin{array}{l}a)\,S = \left\{ {\frac{1}{2};\,2} \right\}\\b)\,\left( {x;\,y} \right) = \left( {1;\,1} \right)\end{array}\)
D.\(\begin{array}{l}a)\,S = \left\{ { - \frac{1}{2};\, - 2} \right\}\\b)\,\left( {x;\,y} \right) = \left( { - 1;\,1} \right)\end{array}\)
\(a) \; \;2{x^2} - 5x + 2 = 0. \) \(b) \; \left \{ \begin{array}{l}2x + y = 1 \ \3x - 2y = 5 \end{array} \right.. \)
A.\(\begin{array}{l}a)\,S = \left\{ { - \frac{1}{2};\, - 2} \right\}\\b)\,\left( {x;\,y} \right) = \left( {1;\, - 1} \right)\end{array}\)
B.\(\begin{array}{l}a)\,S = \left\{ {\frac{1}{2};\,2} \right\}\\b)\,\left( {x;\,y} \right) = \left( {1;\, - 1} \right)\end{array}\)
C.\(\begin{array}{l}a)\,S = \left\{ {\frac{1}{2};\,2} \right\}\\b)\,\left( {x;\,y} \right) = \left( {1;\,1} \right)\end{array}\)
D.\(\begin{array}{l}a)\,S = \left\{ { - \frac{1}{2};\, - 2} \right\}\\b)\,\left( {x;\,y} \right) = \left( { - 1;\,1} \right)\end{array}\)
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Hóa Học lớp 12
