Toán Lớp 9: So sánh
1) 8 và √15 + √17
2) √2020 + √2022 và 2 √2021
3) √2022 – √2021 và √2021 – √2019
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Toán Lớp 9: So sánh
1) 8 và √15 + √17
2) √2020 + √2022 và 2 √2021
3) √2022 – √2021 và √2021 – √2019
Comments ( 1 )
1)\sqrt {15} + \sqrt {17} – 8\\
= \sqrt {17} – 4 – \left( {4 – \sqrt {15} } \right)\\
= \dfrac{{17 – 16}}{{\sqrt {17} + 4}} – \dfrac{{16 – 15}}{{4 + \sqrt {15} }}\\
= \dfrac{1}{{\sqrt {17} + 4}} – \dfrac{1}{{\sqrt {15} + 4}}\\
Do:\sqrt {17} + 4 > \sqrt {15} + 4\\
\Leftrightarrow \dfrac{1}{{\sqrt {17} + 4}} < \dfrac{1}{{\sqrt {15} + 4}}\\
\Leftrightarrow \dfrac{1}{{\sqrt {17} + 4}} – \dfrac{1}{{\sqrt {15} + 4}} < 0\\
\Leftrightarrow \sqrt {15} + \sqrt {17} – 8 < 0\\
\Leftrightarrow \sqrt {15} + \sqrt {17} < 8\\
2)\sqrt {2020} + \sqrt {2022} – 2\sqrt {2021} \\
= \sqrt {2022} – \sqrt {2021} – \left( {\sqrt {2021} – \sqrt {2020} } \right)\\
= \dfrac{1}{{\sqrt {2020} + \sqrt {2021} }} – \dfrac{1}{{\sqrt {2021} + \sqrt {2020} }} < 0\\
\Leftrightarrow \sqrt {2020} + \sqrt {2022} – 2\sqrt {2021} < 0\\
\Leftrightarrow \sqrt {2020} + \sqrt {2022} < 2\sqrt {2021} \\
3)\sqrt {2022} – \sqrt {2021} = \dfrac{1}{{\sqrt {2022} + \sqrt {2021} }}\\
\sqrt {2021} – \sqrt {2019} = \dfrac{2}{{\sqrt {2021} + \sqrt {2019} }} = \dfrac{1}{{\dfrac{{\sqrt {2021} + \sqrt {2019} }}{2}}}\\
Do:\dfrac{1}{{\sqrt {2022} + \sqrt {2021} }} < \dfrac{1}{{\dfrac{{\sqrt {2021} + \sqrt {2019} }}{2}}}\\
\Leftrightarrow \sqrt {2022} – \sqrt {2021} < \sqrt {2021} – \sqrt {2019}
\end{array}$