Toán Lớp 8: cho biếu thức q=(1/x-1-x^2/x^3-1) : x+1/x-1 a, chứng minh rằng q=1/x^2+x+1 b, chưbgs minh rằng qx<1c tìm giá trị lớn nhất của q
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Toán Lớp 8: cho biếu thức q=(1/x-1-x^2/x^3-1) : x+1/x-1 a, chứng minh rằng q=1/x^2+x+1 b, chưbgs minh rằng qx<1c tìm giá trị lớn nhất của q
Comments ( 1 )
a)Dkxd:x \ne 1;x \ne – 1\\
Q = \left( {\dfrac{1}{{x – 1}} – \dfrac{{{x^2}}}{{{x^3} – 1}}} \right):\dfrac{{x + 1}}{{x – 1}}\\
= \left( {\dfrac{1}{{x – 1}} – \dfrac{{{x^2}}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}} \right).\dfrac{{x – 1}}{{x + 1}}\\
= \dfrac{{{x^2} + x + 1 – {x^2}}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{x – 1}}{{x + 1}}\\
= \dfrac{{x + 1}}{{{x^2} + x + 1}}.\dfrac{1}{{x + 1}}\\
= \dfrac{1}{{{x^2} + x + 1}}\\
b)Q.x – 1\\
= \dfrac{1}{{{x^2} + x + 1}}.x – 1\\
= \dfrac{{x – {x^2} – x – 1}}{{{x^2} + x + 1}}\\
= \dfrac{{ – {x^2} – 1}}{{{x^2} + x + 1}}\\
= – \dfrac{{{x^2} + 1}}{{{x^2} + x + 1}}\\
Do:\left\{ \begin{array}{l}
{x^2} + 1 > 0\\
{x^2} + x + 1 > 0
\end{array} \right.\\
\Leftrightarrow \dfrac{{{x^2} + 1}}{{{x^2} + x + 1}} > 0\\
\Leftrightarrow – \dfrac{{{x^2} + 1}}{{{x^2} + x + 1}} < 0\\
\Leftrightarrow Q.x < 1\\
c)Q = \dfrac{1}{{{x^2} + x + 1}}\\
{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}\\
\Leftrightarrow \dfrac{1}{{{x^2} + x + 1}} \le \dfrac{4}{3}\\
\Leftrightarrow Q \le \dfrac{4}{3}\\
\Leftrightarrow GTLN:Q = \dfrac{4}{3}\,khi:x = – \dfrac{1}{2}
\end{array}$