a/ Có: \(\sqrt{99}< \sqrt{100}=10\) mà 10 < 11
=> \(11>\sqrt{99}\)
b/ có: √11 < √16 =4
=> √11 + 1 < 4 + 1 = 5
hay 5 > √11 + 1
c/ Có: √2 > √1 = 1
=> √2 + 1 > 1 + 1 = 2
hay 2 < 1 + √2
d/ 3√11 = √99 ; 12 = √144
mà √99 < √144
=> 3√11 < 12
e/ - 10 = -√100 ; -2√23 = -√92
Có: √100 > √92 => -√100 < - √92
hay -10 < -2√23
f/ Có: √7 < √9 = 3
=> 1 + √7 < 1 + 3 = 4
hay 4 > 1 + √7
Giải:
a) Ta có:
\(11=\sqrt{121}\)
Vì \(\sqrt{121}>\sqrt{99}\)
\(\Leftrightarrow11>\sqrt{99}\)
Vậy ...
b) Ta có:
\(5=4+1=\sqrt{16}+1\)
Vì \(\sqrt{16}+1>\sqrt{11}+1\)
\(\Leftrightarrow5>\sqrt{11}+1\)
Vậy ...
c) Ta có:
\(2=1+1=\sqrt{1}+1\)
Vì \(\sqrt{1}+1< 1+\sqrt{2}\)
\(\Leftrightarrow2< 1+\sqrt{2}\)
Vậy ...
d) Ta có:
\(3\sqrt{11}=\sqrt{9.11}=\sqrt{99}\)
\(12=\sqrt{144}\)
Vì \(\sqrt{99}< \sqrt{144}\)
\(\Leftrightarrow3\sqrt{11}< 12\)
Vậy ...
e) Ta có:
\(-10=-\sqrt{100}\)
\(-2\sqrt{23}=-\sqrt{92}\)
Vì \(-\sqrt{100}< -\sqrt{92}\)
\(\Leftrightarrow-10< -2\sqrt{23}\)
Vậy ...
f) Ta có:
\(4=1+3=1+\sqrt{9}\)
Vì \(1+\sqrt{9}>1+\sqrt{7}\)
\(\Leftrightarrow4>1+\sqrt{7}\)
Vậy ...