Toán Lớp 9: so sánh 5-2 căn 7 và 3 -căn 10
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Toán Lớp 9: so sánh 5-2 căn 7 và 3 -căn 10
Comments ( 1 )
5 – 2\sqrt 7 = \dfrac{{{5^2} – {{\left( {2\sqrt 7 } \right)}^2}}}{{5 + 2\sqrt 7 }} = \dfrac{{ – 3}}{{5 + 2\sqrt 7 }}\\
3 – \sqrt {10} = \dfrac{{9 – 10}}{{3 + \sqrt {10} }} = \dfrac{{ – 1}}{{3 + \sqrt {10} }} = \dfrac{{ – 3}}{{9 + 3\sqrt {10} }}\\
Xet:5 + 2\sqrt 7 – 9 – 3\sqrt {10} \\
= \sqrt {28} – \sqrt {90} – 4 < 0\\
\Leftrightarrow 5 + 2\sqrt 7 < 9 + 3\sqrt {10} \\
\Leftrightarrow \dfrac{1}{{5 + 2\sqrt 7 }} > \dfrac{1}{{9 + 3\sqrt {10} }}\\
\Leftrightarrow \dfrac{{ – 3}}{{5 + 2\sqrt 7 }} < \dfrac{{ – 3}}{{9 + 3\sqrt {10} }}\\
\Leftrightarrow 5 – 2\sqrt 7 < 3 – \sqrt {10} \\
Vậy\,5 – 2\sqrt 7 < 3 – \sqrt {10}
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