so sanh \(\sqrt{8}\)va \(\sqrt{5}\)+1
SO SANH
\(-2\sqrt{5}va-5\sqrt{2}\)
\(-2\sqrt{5}=-\sqrt{2^2.5}=-\sqrt{20}\)
\(-5\sqrt{2}=-\sqrt{5^2.2}=-\sqrt{50}\)
\(\Rightarrow-\sqrt{20}>-\sqrt{50}\)
hay \(-2\sqrt{5}>-5\sqrt{2}\)
\(-2\sqrt{5}=-\sqrt{2^2.5}=-\sqrt{20}\)
\(-5\sqrt{2}=-\sqrt{5^2.2}=-\sqrt{50}\)
vì \(20< 50\)nen \(-\sqrt{20}>-\sqrt{50}\)
so sanh
\(2\sqrt{5}-5\)va \(\sqrt{5}-3\)
\(\sqrt{17}+\sqrt{26}\) va 9
a ) \(2\sqrt{5}-5\) và \(\sqrt{5}-3\)
Ta có ; \(2\sqrt{5}-5-\left(\sqrt{5}-3\right)\)
\(=\sqrt{5}-8\)
\(=\sqrt{5}-\sqrt{64}< 0\)
\(\Rightarrow2\sqrt{5}-5< \sqrt{5}-3\)
Vậy .................
b ) \(\sqrt{17}+\sqrt{26}\) và 9
Ta có :
\(\sqrt{17}>\sqrt{16}\)
\(\sqrt{26}>\sqrt{25}\)
\(\Rightarrow\sqrt{17}+\sqrt{26}>\sqrt{16}+\sqrt{25}=4+5=9\)
Vậy ...
So sanh:
a, \(\sqrt{\dfrac{35}{34}}\) va \(\sqrt{\dfrac{71}{70}}\)
b, \(4\sqrt{5}-3\sqrt{2}\) va 5
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so sanh \(\sqrt{49-16}\) va\(\sqrt{49}\) -\(\sqrt{16}\)
\(\sqrt{0,04}\) va \(0,04\)
\(8\) va \(\sqrt{8}\)
bai 4 so sanh cac so thuc
\(\frac{4}{9}va\)0,4(5)
\(\sqrt[2]{3}va\sqrt[3]{2}\)
So sanh : \(\sqrt{2016}-\sqrt{2015}va\sqrt{\sqrt{2015}-}\sqrt{2014}\)
so sanh \(\sqrt{5}+\sqrt{7}\) va 2\(\sqrt{6}\)
giai full giup minh nhe!thanks
Ta có : \(\left(\sqrt{5}+\sqrt{7}\right)^2=5+7+2\sqrt{35}\)
=\(12+2\sqrt{35}\le12+2\sqrt{36}=12+2.6=24\)
Mà \(\left(2\sqrt{6}\right)^2=24\)
Do đó \(\left(\sqrt{5}+\sqrt{7}\right)^2< \left(2\sqrt{6}\right)^2\)
Mà \(\sqrt{5}+\sqrt{7}>0\) và \(2\sqrt{6}>0\)
Vậy \(\sqrt{5}+\sqrt{7}< 2\sqrt{6}\)
so sanh 4+\(\sqrt{33}va\sqrt{29}+\sqrt{14}\)
Ta có :
\(4+\sqrt{33}>4+\sqrt{25}=4+5=9\)
\(\sqrt{29}+\sqrt{14}< \sqrt{25}+\sqrt{9}=5+3=8\)
Vì \(9>8\) nên \(4+\sqrt{33}>\sqrt{29}+\sqrt{14}\)
Vậy \(4+\sqrt{33}>\sqrt{29}+\sqrt{14}\)
Sorry nhầm !!!! làm tại
\(\sqrt{29}+\sqrt{14}< \sqrt{33}+\sqrt{16}=\sqrt{33}+4\)
Vậy \(\sqrt{33}+4>\sqrt{29}+\sqrt{14}\)
so sanh
\(\sqrt{2020}-\sqrt{2019}va\sqrt{2021}-\sqrt{2020}\)
Ta có: \(\sqrt{2020}-\sqrt{2019}=\frac{\left(\sqrt{2020}-\sqrt{2019}\right)\left(\sqrt{2020}+\sqrt{2019}\right)}{\sqrt{2020}+\sqrt{2019}}\)
\(=\frac{2020-2019}{\sqrt{2020}+\sqrt{2019}}=\frac{1}{\sqrt{2020}+\sqrt{2019}}\)
\(\sqrt{2021}-\sqrt{2020}=\frac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}\)
\(=\frac{2021-2020}{\sqrt{2021}+\sqrt{2020}}=\frac{1}{\sqrt{2021}+\sqrt{2020}}\)
Vì \(\sqrt{2020}+\sqrt{2019}< \sqrt{2021}+\sqrt{2020}\)
\(\Rightarrow\) \(\frac{1}{\sqrt{2020}+\sqrt{2019}}>\frac{1}{\sqrt{2021}+\sqrt{2020}}\)
Hay \(\sqrt{2020}-\sqrt{2019}>\sqrt{2021}-\sqrt{2020}\)
Chúc bn học tốt!