Toán Lớp 8: Cho A= -(x-1)/ x^2+x+1 . Tìm x để A>0, A<0, A=1/3
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Toán Lớp 8: Cho A= -(x-1)/ x^2+x+1 . Tìm x để A>0, A<0, A=1/3
Comments ( 1 )
A = – \dfrac{{x – 1}}{{{x^2} + x + 1}}\\
Xet:{x^2} + x + 1\\
= {x^2} + 2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4} > 0\\
\Leftrightarrow {x^2} + x + 1 > 0\\
+ Khi:A > 0\\
\Leftrightarrow – \dfrac{{x – 1}}{{{x^2} + x + 1}} > 0\\
\Leftrightarrow \dfrac{{x – 1}}{{{x^2} + x + 1}} < 0\\
\Leftrightarrow x – 1 < 0\left( {do:{x^2} + x + 1 > 0} \right)\\
\Leftrightarrow x < 1\\
+ Khi:A < 0\\
\Leftrightarrow – \dfrac{{x – 1}}{{{x^2} + x + 1}} < 0\\
\Leftrightarrow \dfrac{{x – 1}}{{{x^2} + x + 1}} > 0\\
\Leftrightarrow x – 1 > 0\left( {do:{x^2} + x + 1 > 0} \right)\\
\Leftrightarrow x > 1\\
+ Khi:A = \dfrac{1}{3}\\
\Leftrightarrow – \dfrac{{x – 1}}{{{x^2} + x + 1}} = \dfrac{1}{3}\\
\Leftrightarrow – 3x + 3 = {x^2} + x + 1\\
\Leftrightarrow {x^2} + 4x – 2 = 0\\
\Leftrightarrow {x^2} + 4x + 4 = 6\\
\Leftrightarrow {\left( {x + 2} \right)^2} = 6\\
\Leftrightarrow x = – 2 \pm \sqrt 6
\end{array}$