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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
\(\Rightarrow\) \(\sqrt{11}-\sqrt{3}< 2\)
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) \(2\sqrt[3]{3}=\sqrt[3]{2^3}.\sqrt[3]{3}=\sqrt[3]{2^3.3}=\sqrt[3]{24}\)
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 \(33< 3\sqrt[3]{1333}\)
Đặt A = \(\sqrt{ }\)2003 + \(\sqrt{ }\)2005 ; B = 2\(\sqrt{ }\)2004
A² = 2003 + 2005 + 2\(\sqrt{ }\)(2003.2005)
= 4008 + 2\(\sqrt{ }\)[(2004-1)(2004+1)]
= 4008 + 2\(\sqrt{ }\)(2004² - 1) < 2.2004 + 2\(\sqrt{ }\)(2004²) = 4.2004 = B²
\(\Rightarrow\) A < B
Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) (bạn tự c/m) với a = 2003 , b = 2005
được : \(\frac{\sqrt{2003}+\sqrt{2005}}{2}< \sqrt{\frac{2003+2005}{2}}\)
\(\Rightarrow\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)
a)\(\left(\sqrt{2019.2021}\right)^2=2019.2021=\left(2020-1\right)\left(2020+1\right)=2020^2-1< 2020^2\)
=> \(\sqrt{2019.2021}< 2020\)
b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>5+2\sqrt{4}=5+2.2=9\)
=> \(\sqrt{2}+\sqrt{3}>3\)
c) \(9+4\sqrt{5}=4+4\sqrt{5}+5=\left(2+\sqrt{5}\right)^2>\left(2+\sqrt{4}\right)^2=\left(2+2\right)^2=16\)
=> \(9+4\sqrt{5}>16\)
d) \(\sqrt{11}-\sqrt{3}>\sqrt{9}-\sqrt{1}=3-1=2\)
=> \(\sqrt{11}-\sqrt{3}>2\)
a) \(9=6+3=6+\sqrt{9}\)
\(6+2\sqrt{2}=6+\sqrt{8}\)
\(\sqrt{8}< \sqrt{9}\) nên \(6+\sqrt{8}=6+2\sqrt{2}< 6+\sqrt{9}=9\)
b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}=5+\sqrt{24}\)
\(3^2=9=5+4=5+\sqrt{16}\)
\(\sqrt{16}< \sqrt{24}\Rightarrow3^2< \left(\sqrt{2}+\sqrt{3}\right)^2\Rightarrow3< \sqrt{2}+\sqrt{3}\)
c) \(9+4\sqrt{5}=\left(2+\sqrt{5}\right)^2\)
\(16=\left(2+2\right)^2=\left(2+\sqrt{4}\right)^2\)
\(\sqrt{4}< \sqrt{5}\Rightarrow2+\sqrt{4}< 2+\sqrt{5}\Rightarrow\left(2+\sqrt{4}\right)^2=16< \left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)
d) \(\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}=14-\sqrt{132}\)
\(2^2=14-10=14-\sqrt{100}\)
\(\sqrt{100}< \sqrt{132}\Leftrightarrow-\sqrt{100}>-\sqrt{132}\Leftrightarrow14-\sqrt{100}>14-\sqrt{132}\)
\(\Rightarrow2>\sqrt{11}-\sqrt{3}\)