Tìm x X^2-2x-1=0 Tìm GTNN của biểu thức A=3x^2+2x+1 B=10x^2+4y^2-4xy+12x-30 03/10/2021 Bởi Kennedy Tìm x X^2-2x-1=0 Tìm GTNN của biểu thức A=3x^2+2x+1 B=10x^2+4y^2-4xy+12x-30
$$\eqalign{ & {x^2} – 2x – 1 = 0 \cr & \Delta ‘ = 1 + 1 = 2 > 0 \cr & \Rightarrow PT\,\,co\,\,2\,\,nghiem\,\,pb \cr & \Rightarrow \left[ \matrix{ {x_1} = 1 + \sqrt 2 \hfill \cr {x_2} = 1 – \sqrt 2 \hfill \cr} \right. \cr & \cr & A = 3{x^2} + 2x + 1 \cr & A = 3\left( {{x^2} + {2 \over 3}x} \right) + 1 \cr & A = 3\left( {{x^2} + 2x.{1 \over 3} + {1 \over 9}} \right) – {1 \over 3} + 1 \cr & A = 3{\left( {x + {1 \over 3}} \right)^2} + {2 \over 3} \cr & 3{\left( {x + {1 \over 3}} \right)^2} \ge 0 \Rightarrow 3{\left( {x + {1 \over 3}} \right)^2} + {2 \over 3} \ge {2 \over 3} \cr & \Rightarrow A \ge {2 \over 3} \Rightarrow \min A = {2 \over 3} \cr & B = 10{x^2} + 4{y^2} – 4xy + 12x – 30 \cr & B = 9{x^2} + 12x + 4 + {x^2} – 4xy + 4{y^2} – 34 \cr & B = {\left( {3x + 2} \right)^2} + {\left( {x – 2y} \right)^2} – 34 \cr & \Rightarrow B \ge – 34 \cr & \Rightarrow \min B = – 34 \Leftrightarrow \left\{ \matrix{ 3x + 2 = 0 \hfill \cr x – 2y = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = – {2 \over 3} \hfill \cr y = – {1 \over 3} \hfill \cr} \right. \cr} $$ Bình luận
Đa) A=3x^2 – 2x-1 = 3(x^2 – 2x/3 + 1/9) – 1 – 1/3 = 3(x-1/3)^2 – 4/3 ≥ -4/3 Dấu “=” xảy ra <=> x-1/3 = 0 <=> x = 1/3 Vậy Min A = -4/3 <=> x = 1/3 Giải thích các bước giải: Bình luận
$$\eqalign{
& {x^2} – 2x – 1 = 0 \cr
& \Delta ‘ = 1 + 1 = 2 > 0 \cr
& \Rightarrow PT\,\,co\,\,2\,\,nghiem\,\,pb \cr
& \Rightarrow \left[ \matrix{
{x_1} = 1 + \sqrt 2 \hfill \cr
{x_2} = 1 – \sqrt 2 \hfill \cr} \right. \cr
& \cr
& A = 3{x^2} + 2x + 1 \cr
& A = 3\left( {{x^2} + {2 \over 3}x} \right) + 1 \cr
& A = 3\left( {{x^2} + 2x.{1 \over 3} + {1 \over 9}} \right) – {1 \over 3} + 1 \cr
& A = 3{\left( {x + {1 \over 3}} \right)^2} + {2 \over 3} \cr
& 3{\left( {x + {1 \over 3}} \right)^2} \ge 0 \Rightarrow 3{\left( {x + {1 \over 3}} \right)^2} + {2 \over 3} \ge {2 \over 3} \cr
& \Rightarrow A \ge {2 \over 3} \Rightarrow \min A = {2 \over 3} \cr
& B = 10{x^2} + 4{y^2} – 4xy + 12x – 30 \cr
& B = 9{x^2} + 12x + 4 + {x^2} – 4xy + 4{y^2} – 34 \cr
& B = {\left( {3x + 2} \right)^2} + {\left( {x – 2y} \right)^2} – 34 \cr
& \Rightarrow B \ge – 34 \cr
& \Rightarrow \min B = – 34 \Leftrightarrow \left\{ \matrix{
3x + 2 = 0 \hfill \cr
x – 2y = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x = – {2 \over 3} \hfill \cr
y = – {1 \over 3} \hfill \cr} \right. \cr} $$
Đa) A=3x^2 – 2x-1
= 3(x^2 – 2x/3 + 1/9) – 1 – 1/3
= 3(x-1/3)^2 – 4/3 ≥ -4/3
Dấu “=” xảy ra <=> x-1/3 = 0
<=> x = 1/3
Vậy Min A = -4/3 <=> x = 1/3
Giải thích các bước giải: