Toán Lớp 8: Tìm giá trị nhỏ nhất của các thức sau :
9. K= x ² – x +1 ;
10. M = x ² +x + 1 ;
11. N= x ² + x ;
12. L= x ²-x
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Toán Lớp 8: Tìm giá trị nhỏ nhất của các thức sau :
9. K= x ² – x +1 ;
10. M = x ² +x + 1 ;
11. N= x ² + x ;
12. L= x ²-x
Comments ( 1 )
9,\\
{K_{\min }} = \dfrac{3}{4} \Leftrightarrow x = \dfrac{1}{2}\\
10,\\
{M_{\min }} = \dfrac{3}{4} \Leftrightarrow x = – \dfrac{1}{2}\\
11,\\
{N_{\min }} = – \dfrac{1}{4} \Leftrightarrow x = – \dfrac{1}{2}\\
12,\\
{L_{\min }} = – \dfrac{1}{4} \Leftrightarrow x = \dfrac{1}{2}
\end{array}\)
9,\\
K = {x^2} – x + 1\\
= \left( {{x^2} – x + \dfrac{1}{4}} \right) + \dfrac{3}{4}\\
= \left[ {{x^2} – 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right] + \dfrac{3}{4}\\
= {\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
{\left( {x – \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow K = {\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4},\,\,\,\forall x\\
\Rightarrow {K_{\min }} = \dfrac{3}{4} \Leftrightarrow {\left( {x – \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x – \dfrac{1}{2} = 0 \Leftrightarrow x = \dfrac{1}{2}\\
\Rightarrow {K_{\min }} = \dfrac{3}{4} \Leftrightarrow x = \dfrac{1}{2}\\
10,\\
M = {x^2} + x + 1\\
= \left( {{x^2} + x + \dfrac{1}{4}} \right) + \dfrac{3}{4}\\
= \left[ {{x^2} + 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right] + \dfrac{3}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
{\left( {x + \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow M = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4},\,\,\,\forall x\\
\Rightarrow {M_{\min }} = \dfrac{3}{4} \Leftrightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x + \dfrac{1}{2} = 0 \Leftrightarrow x = – \dfrac{1}{2}\\
\Rightarrow {M_{\min }} = \dfrac{3}{4} \Leftrightarrow x = – \dfrac{1}{2}\\
11,\\
N = {x^2} + x\\
= \left( {{x^2} + x + \dfrac{1}{4}} \right) – \dfrac{1}{4}\\
= \left[ {{x^2} + 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right] – \dfrac{1}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} – \dfrac{1}{4}\\
{\left( {x + \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow N = {\left( {x + \dfrac{1}{2}} \right)^2} – \dfrac{1}{4} \ge – \dfrac{1}{4},\,\,\,\forall x\\
\Rightarrow {N_{\min }} = – \dfrac{1}{4} \Leftrightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x + \dfrac{1}{2} = 0 \Leftrightarrow x = – \dfrac{1}{2}\\
\Rightarrow {N_{\min }} = – \dfrac{1}{4} \Leftrightarrow x = – \dfrac{1}{2}\\
12,\\
L = {x^2} – x\\
= \left( {{x^2} – x + \dfrac{1}{4}} \right) – \dfrac{1}{4}\\
= \left[ {{x^2} – 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right] – \dfrac{1}{4}\\
= {\left( {x – \dfrac{1}{2}} \right)^2} – \dfrac{1}{4}\\
{\left( {x – \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow L = {\left( {x – \dfrac{1}{2}} \right)^2} – \dfrac{1}{4} \ge – \dfrac{1}{4},\,\,\,\forall x\\
\Rightarrow {L_{\min }} = – \dfrac{1}{4} \Leftrightarrow {\left( {x – \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x – \dfrac{1}{2} = 0 \Leftrightarrow x = \dfrac{1}{2}\\
\Rightarrow {L_{\min }} = – \dfrac{1}{4} \Leftrightarrow x = \dfrac{1}{2}
\end{array}\)