Toán Lớp 9: cho A=(x+√x+10/x-9 – 1/√x-3）/1/√x-3
a) rút gọn A
b)tính A khi x=4
c)Tìm gtnn của A(cô si)
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Toán Lớp 9: cho A=(x+√x+10/x-9 – 1/√x-3）/1/√x-3
a) rút gọn A
b)tính A khi x=4
c)Tìm gtnn của A(cô si)
Comments ( 1 )
a)Dkxd:x \ge 0;x \ne 9\\
A = \left( {\dfrac{{x + \sqrt x + 10}}{{x – 9}} – \dfrac{1}{{\sqrt x – 3}}} \right):\dfrac{1}{{\sqrt x – 3}}\\
= \dfrac{{x + \sqrt x + 10 – \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}.\left( {\sqrt x – 3} \right)\\
= \dfrac{{x + \sqrt x + 10 – \sqrt x – 3}}{{\sqrt x + 3}}\\
= \dfrac{{x + 7}}{{\sqrt x + 3}}\\
b)x = 4\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow A = \dfrac{{4 + 7}}{{2 + 3}} = \dfrac{{11}}{5}\\
c)A = \dfrac{{x + 7}}{{\sqrt x + 3}}\\
= \dfrac{{x – 9 + 16}}{{\sqrt x + 3}}\\
= \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right) + 16}}{{\sqrt x + 3}}\\
= \sqrt x – 3 + \dfrac{{16}}{{\sqrt x + 3}}\\
= \sqrt x + 3 + \dfrac{{16}}{{\sqrt x + 3}} – 6\\
\ge 2\sqrt {\left( {\sqrt x + 3} \right).\dfrac{{16}}{{\sqrt x + 3}}} – 6\\
\Leftrightarrow A \ge 2\sqrt {16} – 6\\
\Leftrightarrow A \ge 2\\
\Leftrightarrow GTNN:A = 2\\
Khi:\left( {\sqrt x + 3} \right) = \dfrac{{16}}{{\sqrt x + 3}}\\
\Leftrightarrow \sqrt x + 3 = 4\\
\Leftrightarrow x = 1\left( {tmdk} \right)
\end{array}$