Bài tập 5.18 trang 202 SBT Toán 11
Bài tập 5.18 trang 202 SBT Toán 11
Rút gọn \({f\left( x \right) = \left[ {\frac{{x - 1}}{{2(\sqrt x + 1)}} + 1} \right].\frac{2}{{\sqrt x + 1}}:{{\left( {\frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} + \sqrt {x - 2} }} + \frac{{x - 2}}{{\sqrt {{x^2} - 4} - x + 2}}} \right)}^2}}\) và tìm
\(\begin{array}{l}
f\left( x \right) = \left[ {\frac{{x - 1}}{{2(\sqrt x + 1)}} + 1} \right].\frac{2}{{\sqrt x + 1}}:{\left( {\frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} + \sqrt {x - 2} }} + \frac{{x - 2}}{{\sqrt {{x^2} - 4} - x + 2}}} \right)^2}\\
= \frac{{x + 2\sqrt x + 1}}{{2\left( {\sqrt x + 1} \right)}}.\frac{2}{{\sqrt x + 1}}:{\left[ {\frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} + \sqrt {x - 2} }} + \frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} - \sqrt {x - 2} }}} \right]^2}\\
= 1:{\left( {\frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} + \sqrt {x - 2} }} + \frac{{\sqrt {x - 2} }}{{\sqrt {x + 2} - \sqrt {x - 2} }}} \right)^2}\\
= 1:{\left[ {\frac{{\sqrt {{x^2} - 4} - \left( {x - 2} \right) + \left( {x - 2} \right) + \sqrt {{x^2} - 4} }}{4}} \right]^2}\\
= 1:{\left( {\frac{{\sqrt {{x^2} - 4} }}{2}} \right)^2} = \frac{4}{{{x^2} - 4}}
\end{array}\)
Suy ra \(f'\left( x \right) = {\left( {\frac{4}{{{x^2} - 4}}} \right)^\prime } = - \frac{{4\left( {2x} \right)}}{{{{\left( {{x^2} - 4} \right)}^2}}} = - \frac{{8x}}{{{{\left( {{x^2} - 4} \right)}^2}}}\)
-- Mod Toán 11
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