so sanh
a) 2^90 va 5^36
b) 99^200 va 9999^100
c) 2^150 va 3^100
d) \(\sqrt{26+2}\) va \(\sqrt{26}\)+ \(\sqrt{2}\)
Những câu hỏi liên quan
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 ...
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a) So sanh: \(\sqrt{17}+\sqrt{26}+1\)va \(\sqrt{99}\)
b) CMR: \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
a)\(\sqrt{17}+\sqrt{26}+1>\sqrt{16}+\sqrt{25}+1=4+5+1=10\)
b) \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+.......+\frac{1}{\sqrt{100}}=\frac{100}{\sqrt{100}}=10\)
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So Sánh \(\sqrt{17}\)+\(\sqrt{26}\)+1 va\(\sqrt{99}\)
\(\sqrt{17}+\sqrt{26}+1>\sqrt{16}+\sqrt{25}+1=4+5+1=10=\sqrt{100}>\sqrt{99}\)
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So sánh:1,2-sqrt{2}vafrac{1}{2}2, 2sqrt{3}-5vasqrt{3}-43, sqrt{3}-3sqrt{2}va-4sqrt{3}+5sqrt{2}4,1-sqrt{3}vasqrt{2}-sqrt{6}5, sqrt{4sqrt{5}}vasqrt{5sqrt{3}}6, sqrt{sqrt{6}-sqrt{5}}-sqrt{sqrt{3}-sqrt{2}}va..07, -2sqrt{frac{1}{2}sqrt{5}}va-3sqrt{frac{1}{3}sqrt{2}}
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So sánh:
1,\(2-\sqrt{2}va\frac{1}{2}\)
2, \(2\sqrt{3}-5va\sqrt{3}-4\)
3, \(\sqrt{3}-3\sqrt{2}va-4\sqrt{3}+5\sqrt{2}\)
4,\(1-\sqrt{3}va\sqrt{2}-\sqrt{6}\)
5, \(\sqrt{4\sqrt{5}}va\sqrt{5\sqrt{3}}\)
6, \(\sqrt{\sqrt{6}-\sqrt{5}}-\sqrt{\sqrt{3}-\sqrt{2}}va..0\)
7, \(-2\sqrt{\frac{1}{2}\sqrt{5}}va-3\sqrt{\frac{1}{3}\sqrt{2}}\)
so sanh
a, \(\left|-2\right|^{300}\) va \(\left|-4\right|^{150}\)
b,\(\left|-2\right|^{300}\)va \(\left|-2\right|^{300}\)
co loi giai
a, Ta có:
\(\left|-2\right|^{300}=2^{300}\) (1)
\(\left|-4\right|^{150}=4^{150}=\left(2^2\right)^{150}=2^{300}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\left|-2\right|^{300}=\left|-4\right|^{150}\)
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a: \(\left|-2\right|^{300}=2^{300}\)
\(\left|-4\right|^{150}=4^{150}=2^{300}\)
Do đó: \(\left|-2\right|^{300}=\left|-4\right|^{150}\)
b: \(\left|-2\right|^{300}=\left|-2\right|^{300}\)
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left(5+4sqrt{2}right)left(3+2sqrt{1+sqrt{2}}right)left(3-2sqrt{1+sqrt{2}}right) sqrt{frac{9}{4}-sqrt{2}} Sosanh2sqrt{27}vasqrt{147}
2sqrt{15}vasqrt{59}
2sqrt{2}-1va2
frac{sqrt{3}}{2}va1
-frac{sqrt{10}}{2}va-2sqrt{5}
sqrt{6}-1va3
2sqrt{5}-5sqrt{2}va1
frac{sqrt{8}}{3}vafrac{3}{4}
-2sqrt{6}va-sqrt{23}
2sqrt{6}-2va3
sqrt{111}-7va4
Xếp theo thứ tự tăng dần: 21,2sqrt{7},15sqrt{3},-sqrt{123} ; 28sqrt{2},sqrt{14},2sqrt{147},36sqrt{4}
giảm dần: 6sqrt{frac{1}{4}},4sqrt{frac{1}{2}...
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\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\\ \\ \\ \sqrt{\frac{9}{4}-\sqrt{2}}\\ \\ \\ Sosanh2\sqrt{27}va\sqrt{147}\\ \\ \\ 2\sqrt{15}va\sqrt{59}\\ \\ \\ 2\sqrt{2}-1va2\\ \\ \\ \frac{\sqrt{3}}{2}va1\\ \\ \\ -\frac{\sqrt{10}}{2}va-2\sqrt{5}\\ \\ \\ \sqrt{6}-1va3\\ \\ \\ 2\sqrt{5}-5\sqrt{2}va1\\ \\ \\ \frac{\sqrt{8}}{3}va\frac{3}{4}\\ \\ \\ -2\sqrt{6}va-\sqrt{23}\\ \\ \\ 2\sqrt{6}-2va3\\ \\ \\ \sqrt{111}-7va4\)
Xếp theo thứ tự tăng dần: \(21,2\sqrt{7},15\sqrt{3},-\sqrt{123}\) ; \(28\sqrt{2},\sqrt{14},2\sqrt{147},36\sqrt{4}\)
giảm dần: \(6\sqrt{\frac{1}{4}},4\sqrt{\frac{1}{2}},-\sqrt{132},2\sqrt{3},\sqrt{\frac{15}{5}}\); \(-27,4\sqrt{3},16\sqrt{5},21\sqrt{2}\)
a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)
=-7
b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)
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So sánh:
1) \(2\sqrt{27}\) và \(\sqrt{147}\)
+ \(2\sqrt{27}\) = \(6\sqrt{3}\)
+ \(\sqrt{147}\) = \(7\sqrt{3}\)
⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)
Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)
2) \(2\sqrt{15}\) và \(\sqrt{59}\)
+ \(2\sqrt{15}\) = \(\sqrt{60}\)
⇒ \(\sqrt{60}\) > \(\sqrt{59}\)
Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)
3) \(2\sqrt{2}-1\) và 2
\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)
So sánh: 3 và \(2\sqrt{2}\)
+ 3 = \(\sqrt{9}\)
+ \(2\sqrt{2}=\sqrt{8}\)
⇒ \(\sqrt{8}\) < \(\sqrt{9}\)
⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1
⇒ \(2\sqrt{2}\) - 1 < 3 - 1
Vậy: \(2\sqrt{2}-1< 2\)
4) \(\frac{\sqrt{3}}{2}\) và 1
+ 1 = \(\frac{2}{2}\)
⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)
Vậy: \(\frac{\sqrt{3}}{2}\) < 1
5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)
+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)
⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)
Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)
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so sanh x va y biet
a) x=\(2\sqrt{7}\)va y=\(3\sqrt{3}\)
b) x=\(6\sqrt{2}\)va y=\(5\sqrt{3}\)
c) x=\(\sqrt{31}-\sqrt{33}\) va y=\(6-\sqrt{11}\)
bai 4 so sanh cac so thuc
\(\frac{4}{9}va\)0,4(5)
\(\sqrt[2]{3}va\sqrt[3]{2}\)
so sánh:
a) \(2^{20}\)và \(5^{36}\)
b) \(99^{200}\)và \(9999^{100}\)
c) \(2^{150}\)và \(3^{100}\)
d) \(\sqrt{26+2}\)và \(\sqrt{26}\)+\(\sqrt{2}\)
a) Ta thấy số dưới lẫn số mũ của 536 lớn hơn 220 => 536>220
b)Ta có:\(99^{200}=99^{100}.99^{100}\)
\(9999^{100}=\left(99.101\right)^{100}=99^{100}.101^{100}\)
VÌ \(99^{100}.99^{100}< 99^{100}.101^{100}\)
Nên: \(99^{200}< 9999^{100}\)
c)Ta có: \(2^{150}=\left(2^3\right)^{50}=8^{50}\)
\(3^{100}=\left(3^2\right)^{50}=9^{50}\)
Vì \(8^{50}< 9^{50}\)nên : \(2^{150}< 3^{100}\)
d)\(\sqrt{26+2}=\sqrt{28}=5< x< 6\)
\(\sqrt{26}+\sqrt{2}=5< x< 6+1< y< 2\)
=> \(\sqrt{26+2}< \sqrt{26}+\sqrt{2}\)
Câu d mình l
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