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a, \(1< 2\Rightarrow\sqrt{1}< \sqrt{2}\Rightarrow1+1< \sqrt{2}+1\Rightarrow2< \sqrt{2}+1\)
c, \(4>3=>\sqrt{4}>\sqrt{3}=>\sqrt{4}-1>\sqrt{3}-1\Rightarrow1>\sqrt{3}-1\)
d, \(16>11=>\sqrt{16}>\sqrt{11}\Rightarrow4>\sqrt{11}=>4.\left(-3\right)< \sqrt{11}.\left(-3\right)\)
\(=>-12< -3.\sqrt{11}\)
a)2=1+1
Có:12<\(\sqrt{2}^{^{ }2}\)
=> 1<\(\sqrt{2}\)
=>1+1<\(\sqrt{2}+1\)
=>2<\(\sqrt{2}+1\)
c) 10=2.5
Có;\(5=\)\(\sqrt{25}< \sqrt{31}\)
=>\(\sqrt{31}>\sqrt{25}\)
=>\(2.\sqrt{31}>2.\sqrt{25}\)
=>\(2.\sqrt{31}>10\)
b) 1=2-1
Có: \(2=\sqrt{4}>\sqrt{3}\)
=>\(\sqrt{4}-1>\sqrt{3}-1\)
=>\(1>\sqrt{3}-1\)
d) -12=-3.4
Có:\(4=\sqrt{16}>\sqrt{11}\)
=>\(\sqrt{11}< \sqrt{16}\)
=>\(-3.\sqrt{11}>-3.\sqrt{16}\)
=>\(-3.\sqrt{11}>-12\)
a. Ta có : \(\sqrt{8}< \sqrt{9}\) ( vì 8< 9)
hay \(2\sqrt{2}< 3\)
\(\Rightarrow\) \(2\sqrt{2}+6< 3+6\)
hay \(2\sqrt{2}+6< 9\)
b. Ta có : \(\sqrt{6}>\sqrt{4}\) (vì 6 > 4 )
hay \(\sqrt{2.3}>2\)
\(\Rightarrow\) 2\(\sqrt{2.3}\) > 4
\(\Rightarrow\) 2 + \(2\sqrt{2.3}\) + 3 > 9
hay \(\left(\sqrt{2}+\sqrt{3}\right)^2\)> 9
\(\Rightarrow\) \(\sqrt{2}+\sqrt{3}>3\)
c. Ta có: \(\sqrt{80}>\sqrt{49}\) (vì 80>49)
hay \(4\sqrt{5}\) > 7
\(\Rightarrow\) 9 + \(4\sqrt{5}\) > 16
d. Ta có : \(2\sqrt{33}>2\sqrt{25}\) (vì 33> 25 ) hay \(2\sqrt{23}>2.5\)
\(\Rightarrow\) - \(2\sqrt{33}\) < - 2.5
\(\Rightarrow\) 11 - \(2\sqrt{11.3}\) +3 < 11- 2.5 +3
hay \(\left(\sqrt{11}-\sqrt{3}\right)^2\) < 4
\(\Rightarrow\) \(\sqrt{11}-\sqrt{3}< 2\)
Ta có :\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{25}}\left(1\right);\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{25}}\left(2\right);\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{25}}\left(3\right);...;\frac{1}{\sqrt{24}}>\frac{1}{\sqrt{25}}\left(24\right);\frac{1}{\sqrt{25}}=\frac{1}{\sqrt{25}}\left(25\right)\)
Cộng các vế từ (1) -> (25),ta có :\(A>\frac{1}{\sqrt{25}}.25=\frac{25}{5}=5\)
P/S : Theo cách làm trên,ta có công thức tổng quát :\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{n}}>\sqrt{n}\left(n\in N;n>1\right)\)
a/ \(\left(\sqrt{2}+\sqrt{3}\right)^2=2+3+2\sqrt{2.3}=5+2\sqrt{6}=5+\sqrt{24}\)
\(\left(\sqrt{10}\right)^2=10=5+5=5+\sqrt{25}\)
Vì \(\sqrt{24}< \sqrt{25}\)
=>\(\sqrt{2}+\sqrt{3}< \sqrt{10}\)
b/\(\left(\sqrt{3}+2\right)^2=3+4+4\sqrt{3}=7+4\sqrt{3}\)
\(\left(\sqrt{2}+\sqrt{16}\right)^2=2+16+2\sqrt{2.16}=18+4\sqrt{8}\)
=> \(\sqrt{3}+2< \sqrt{2}+\sqrt{16}\)
c/ \(16=\sqrt{16^2}\)
\(\sqrt{15}.\sqrt{17}=\sqrt{15.17}=\sqrt{\left(16-1\right)\left(16+1\right)}=\sqrt{16^2-1}\)
=> \(16>\sqrt{15}.\sqrt{17}\)
d/\(8^2=64=32+32=32+2\sqrt{256}\)
\(\left(\sqrt{15}+\sqrt{17}\right)^2=15+17+2\sqrt{15.17}=32+2\sqrt{255}\)
=> \(8>\sqrt{15}+\sqrt{17}\)
a) \(2\sqrt[3]{3}=\sqrt[3]{2^3}.\sqrt[3]{3}=\sqrt[3]{2^3.3}=\sqrt[3]{24}\)
Ta có : \(24>23\), nên \(\sqrt[3]{24}>\sqrt[3]{23}\)
Vậy \(2\sqrt[3]{3}>\sqrt[3]{23}\)
b) Ta có :
\(11=\sqrt[3]{11^3}=\sqrt[3]{1331}\)
Từ đó suy ra \(33< 3\sqrt[3]{1333}\)