Toán Lớp 8: Cho biểu thức B=(x-1)/2:((x^2+2)/(x^3-1)+x/(x^2+x+1)+1/(1-x))
a) Chứng minh biểu thức B > 0 ∀ x≠1
b) Tìm GTNN của B
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Toán Lớp 8: Cho biểu thức B=(x-1)/2:((x^2+2)/(x^3-1)+x/(x^2+x+1)+1/(1-x))
a) Chứng minh biểu thức B > 0 ∀ x≠1
b) Tìm GTNN của B
Comments ( 1 )
a)B > 0\\
b)Min = \dfrac{3}{8}
\end{array}\)
a)B = \dfrac{{x – 1}}{2}:\left( {\dfrac{{{x^2} + 2}}{{{x^3} – 1}} + \dfrac{x}{{{x^2} + x + 1}} + \dfrac{1}{{1 – x}}} \right)\\
= \dfrac{{x – 1}}{2}:\dfrac{{{x^2} + 2 + x\left( {x – 1} \right) – {x^2} – x – 1}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{x – 1}}{2}.\dfrac{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}{{1 + {x^2} – x – x}}\\
= \dfrac{{x – 1}}{2}.\dfrac{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}{{{{\left( {x – 1} \right)}^2}}}\\
= \dfrac{{{x^2} + x + 1}}{2}\\
Do:{x^2} + x + 1 = {x^2} + 2x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
Do:{\left( {x + \dfrac{1}{2}} \right)^2} \ge 0\forall x\\
\to {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\\
\to {x^2} + x + 1 > 0\\
\to \dfrac{{{x^2} + x + 1}}{2} > 0\\
\to B > 0\\
\to dpcm\\
b)B = \dfrac{{{x^2} + x + 1}}{2} = \dfrac{{{{\left( {x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}}{2}\\
Do:{\left( {x + \dfrac{1}{2}} \right)^2} \ge 0\forall x\\
\to {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\to \dfrac{{{{\left( {x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}}{2} \ge \dfrac{3}{8}\\
\to Min = \dfrac{3}{8}\\
\Leftrightarrow x = – \dfrac{1}{2}
\end{array}\)