Toán Lớp 9: cho A = cănx/cănx – 1 + 2/x – cănx và B = 1/cănx – 1 (x>0; x#1)
a) tính B khi x = 9
b) C = A:B rút gọn C
c) x? C=3
d) so sánh C với 1/4
e) cm C>2
f) x thuộc Z để C thuộc Z
g) min C
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Toán Lớp 9: cho A = cănx/cănx – 1 + 2/x – cănx và B = 1/cănx – 1 (x>0; x#1)
a) tính B khi x = 9
b) C = A:B rút gọn C
c) x? C=3
d) so sánh C với 1/4
e) cm C>2
f) x thuộc Z để C thuộc Z
g) min C
Comments ( 1 )
a)x = 9\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 3\\
B = \dfrac{1}{{\sqrt x – 1}} = \dfrac{1}{{3 – 1}} = \dfrac{1}{2}\\
b)C = A:B\\
= \left( {\dfrac{{\sqrt x }}{{\sqrt x – 1}} + \dfrac{2}{{x – \sqrt x }}} \right):\dfrac{1}{{\sqrt x – 1}}\\
= \dfrac{{\sqrt x .\sqrt x + 2}}{{\sqrt x \left( {\sqrt x – 1} \right)}}.\left( {\sqrt x – 1} \right)\\
= \dfrac{{x + 2}}{{\sqrt x }}\\
c)C = 3\\
\Leftrightarrow \dfrac{{x + 2}}{{\sqrt x }} = 3\\
\Leftrightarrow x + 2 = 3\sqrt x \\
\Leftrightarrow x – 3\sqrt x + 2 = 0\\
\Leftrightarrow \left( {\sqrt x – 1} \right)\left( {\sqrt x – 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = 1 \Leftrightarrow x = 1\left( {ktm} \right)\\
\sqrt x = 2 \Leftrightarrow x = 4\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 4\\
d)C – \dfrac{1}{4}\\
= \dfrac{{x + 2}}{{\sqrt x }} – \dfrac{1}{4}\\
= \dfrac{{4x + 8 – \sqrt x }}{{4\sqrt x }}\\
= \dfrac{{4x – \sqrt x + 8}}{{4\sqrt x }}\\
= \dfrac{{{{\left( {2\sqrt x } \right)}^2} – 2.2\sqrt x .\dfrac{1}{4} + \dfrac{1}{{16}} + \dfrac{{127}}{{16}}}}{{4\sqrt x }}\\
= \dfrac{{{{\left( {2\sqrt x – \dfrac{1}{4}} \right)}^2} + \dfrac{{127}}{{16}}}}{{4\sqrt x }} > 0\\
\Leftrightarrow C > \dfrac{1}{4}\\
e)C – 2\\
= \dfrac{{x + 2}}{{\sqrt x }} – 2\\
= \dfrac{{x + 2 – 2\sqrt x }}{{\sqrt x }}\\
= \dfrac{{x – 2\sqrt x + 1 + 1}}{{\sqrt x }}\\
= \dfrac{{{{\left( {\sqrt x – 1} \right)}^2} + 1}}{{\sqrt x }} > 0\\
\Leftrightarrow C > 2\\
f)C = \dfrac{{x + 2}}{{\sqrt x }} = \sqrt x + \dfrac{2}{{\sqrt x }}\\
\Leftrightarrow \sqrt x \in U\left( 2 \right)\\
\Leftrightarrow \sqrt x = 2\left( {do:\sqrt x \# 1} \right)\\
\Leftrightarrow x = 4\left( {tm} \right)\\
Vậy\,x = 4\\
g)C = \dfrac{{x + 2}}{{\sqrt x }} = \sqrt x + \dfrac{2}{{\sqrt x }}\\
Theo\,Co – si:\\
\sqrt x + \dfrac{2}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{2}{{\sqrt x }}} = 2\sqrt 2 \\
C \ge 2\sqrt 2 \\
\Leftrightarrow GTNN:C = 2\sqrt 2 \,khi:x = 2
\end{array}$