2 câu trả lời
Đáp án:
Giải thích các bước giải: A=3x2+2y2va4x+y=1Taco:4x+y=1⇒y=1−4x⇒A=3x2+2y2=3x2+2.(1−4x)2=3x2+2.(1−8x+16x2)=3x2+2−16x+32x2=35x2−16x+2=35.(x2−1635x)+2=35.(x2−2.835.x+(835)2)+2−35.(835)2=35.(x−835)2+635⇒A≥635⇒Amin
\eqalign{ & A = 3{x^2} + 2{y^2} \cr & 4x + y = 1 \Leftrightarrow y = 1 - 4x \cr & \Rightarrow A = 3{x^2} + 2{\left( {1 - 4x} \right)^2} \cr & A = 3{x^2} + 2\left( {1 - 8x + 16{x^2}} \right) \cr & A = 35{x^2} - 16x + 2 \cr & A = 35\left( {{x^2} - {{16} \over {35}}x} \right) + 2 \cr & A = 35\left( {{x^2} - 2.x.{8 \over {35}} + {{\left( {{8 \over {35}}} \right)}^2}} \right) - 35{\left( {{8 \over {35}}} \right)^2} + 2 \cr & A = 35{\left( {x - {8 \over {35}}} \right)^2} + {6 \over {35}} \cr & {\left( {x - {8 \over {35}}} \right)^2} \ge 0 \Rightarrow 35{\left( {x - {8 \over {35}}} \right)^2} \ge 0 \Leftrightarrow 35{\left( {x - {8 \over {35}}} \right)^2} + {6 \over {35}} \ge {6 \over {35}} \cr & \Rightarrow A \ge {6 \over {35}} \cr & \Rightarrow {A_{\min }} = {6 \over {35}} \Leftrightarrow x = {8 \over {35}} \cr}