Toán Lớp 9: so sanh
a,5$\sqrt[]{3}$-4 và 3$\sqrt[]{5}$
b, $\sqrt{26}$+$\sqrt{17}$+$\sqrt{10}$+$\sqrt{5}$+1 và 15
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Toán Lớp 9: so sanh
a,5$\sqrt[]{3}$-4 và 3$\sqrt[]{5}$
b, $\sqrt{26}$+$\sqrt{17}$+$\sqrt{10}$+$\sqrt{5}$+1 và 15
Comments ( 1 )
a,\\
5\sqrt 3 – 4 < 3\sqrt 5 \\
b,\\
\sqrt {26} + \sqrt {17} + \sqrt {10} + \sqrt 5 + 1 > 15
\end{array}\)
a,\\
5\sqrt 3 – 4 = \sqrt {{5^2}.3} – 4 = \sqrt {75} – 4 < \sqrt {81} – 4 = \sqrt {{9^2}} – 4\\
= 9 – 4 = 5 < 6 = 3.2 = 3.\sqrt 4 < 3.\sqrt 5 \\
\Rightarrow 5\sqrt 3 – 4 < 3\sqrt 5 \\
b,\\
\sqrt {26} + \sqrt {17} + \sqrt {10} + \sqrt 5 + 1\\
> \sqrt {25} + \sqrt {16} + \sqrt 9 + \sqrt 4 + 1\\
= \sqrt {{5^2}} + \sqrt {{4^2}} + \sqrt {{3^2}} + \sqrt {{2^2}} + 1\\
= 5 + 4 + 3 + 2 + 1\\
= 15\\
\Rightarrow \sqrt {26} + \sqrt {17} + \sqrt {10} + \sqrt 5 + 1 > 15
\end{array}\)