Toán Lớp 7: cho các đa thức:
f(x) = 3x^2 – 18x + 27; g(x)=/x-3/ ; h(x)= f(x) – 2g(x) +1
Tìm x để h(x) đạt giá trị nhỏ nhất
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Toán Lớp 7: cho các đa thức:
f(x) = 3x^2 – 18x + 27; g(x)=/x-3/ ; h(x)= f(x) – 2g(x) +1
Tìm x để h(x) đạt giá trị nhỏ nhất
Comments ( 1 )
f\left( x \right) = 3{x^2} – 18x + 27\\
g\left( x \right) = \left| {x – 3} \right|\\
h\left( x \right) = f\left( x \right) – 2g\left( x \right) + 1\\
= 3{x^2} – 18x + 27 – 2\left| {x – 3} \right| + 1\\
= 3\left( {{x^2} – 6x + 9} \right) – 2\left| {x – 3} \right| + 1\\
= 3.{\left( {x – 3} \right)^2} – 2\left| {x – 3} \right| + 1\\
= 3.{\left( {\left| {x – 3} \right|} \right)^2} – 2\left| {x – 3} \right| + 1\\
3.\left[ {{{\left( {\left| {x – 3} \right|} \right)}^2} – \dfrac{2}{3}\left| {x – 3} \right|} \right] + 1\\
= 3.\left[ {{{\left( {\left| {x – 3} \right|} \right)}^2} – 2.\left| {x – 3} \right|.\dfrac{1}{3} + \dfrac{1}{9}} \right] – 3.\dfrac{1}{9} + 1\\
= 3.{\left( {\left| {x – 3} \right| – \dfrac{1}{3}} \right)^2} + \dfrac{2}{3} \ge \dfrac{2}{3}\\
\Leftrightarrow h\left( x \right) \ge \dfrac{2}{3}\\
\Leftrightarrow GTNN:h\left( x \right) = \dfrac{2}{3}\,\\
Khi:\left| {x – 3} \right| = \dfrac{1}{3} \Leftrightarrow \left[ \begin{array}{l}
x – 3 = \dfrac{1}{3} \Leftrightarrow x = \dfrac{{10}}{3}\\
x – 3 = \dfrac{{ – 1}}{3} \Leftrightarrow x = \dfrac{8}{3}
\end{array} \right.\\
Vay\,x = \dfrac{{10}}{3};x = \dfrac{8}{3}
\end{array}$