Giải hệ phương trình: 5a^2 + 3b^2 = 23; a + b = 1
Câu hỏi:
Giải hệ phương trình: \[\left\{ {\begin{array}{*{20}{c}}{5{a^2} + 3{b^2} = 23}\\{a + b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\]
Trả lời:
\[\left\{ {\begin{array}{*{20}{c}}{5{a^2} + 3{b^2} = 23}\\{a + b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{5{a^2} + 3{{(1 - a)}^2} = 23}\\{b = 1 - a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\]
\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{5{a^2} + 3{a^2} - 6a + 3 = 23}\\{b = 1 - a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{8{a^2} - 6a - 20 = 0}\\{b = 1 - a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\]
\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{(a - 2)(4a + 5) = 0}\\{b = 1 - a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\]
\[ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{a = 2}\\{b = - 1}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{a = \frac{{ - 5}}{4}}\\{b = \frac{9}{4}}\end{array}} \right.}\end{array}} \right.\]
Vậy các nghiệm (a; b) của hệ phương trình là: (2; –1) và \[\left( {\frac{{ - 5}}{4};\,\frac{9}{4}} \right)\].
