Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}A = 3{x^2} + 2{y^2}\,\,\,\,\,\,\,\,va\,\,\,\,\,\,\,4x + y = 1\\Ta\,co:\\4x + y = 1 \Rightarrow y = 1 - 4x\\ \Rightarrow A = 3{x^2} + 2{y^2}\\ = 3{x^2} + 2.{\left( {1 - 4x} \right)^2}\\ = 3{x^2} + 2.\left( {1 - 8x + 16{x^2}} \right)\\ = 3{x^2} + 2 - 16x + 32{x^2}\\ = 35{x^2} - 16x + 2\\ = 35.\left( {{x^2} - \frac{{16}}{{35}}x} \right) + 2\\ = 35.\left( {{x^2} - 2.\frac{8}{{35}}.x + {{\left( {\frac{8}{{35}}} \right)}^2}} \right) + 2 - 35.{\left( {\frac{8}{{35}}} \right)^2}\\ = 35.{\left( {x - \frac{8}{{35}}} \right)^2} + \frac{6}{{35}}\\ \Rightarrow A \ge \frac{6}{{35}}\\ \Rightarrow {A_{\min }} = \frac{6}{{35}} \Leftrightarrow x = \frac{8}{{35}} \Rightarrow y = 1 - 4.\frac{8}{{35}} = \frac{3}{{35}}\end{array}\)
