a) \(1=\sqrt{1}< \sqrt{2}\)

b) \(2=\sqrt{4}>\sqrt{3}\)

c) \(6=\sqrt{36}< \sqrt{41}\)

d) \(7=\sqrt{49}>\sqrt{47}\)

e) \(2=1+1=\sqrt{1}+1< \sqrt{2}+1\)

f) \(1=2-1=\sqrt{4}-1>\sqrt{3}-1\)

g) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\)

h) \(\sqrt{3}>0>-\sqrt{12}\)

i) \(5=\sqrt{25}< \sqrt{29}\)

\(\Rightarrow-5>-\sqrt{29}\)

b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)

mà 80>75

nên \(4\sqrt{5}>5\sqrt{3}\)

a: ĐKXĐ: \(x>0\)

b: Ta có: \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)

\(=x+\sqrt{x}-2\sqrt{x}-1+1\)

\(=x-\sqrt{x}\)

\(2^{35}< 2^{36}=8^{12}\)

\(3^{24}=\left(3^2\right)^{12}=9^{12}\)

\(8^{12}< 9^{12}\)

=>\(2^{36}< 3^{24}\)

=>\(2^{35}< 3^{24}\)

a) \(\sqrt{2017}-2\sqrt{2016}=\sqrt{2017}-\sqrt{8064}< 0< \sqrt{2016}\)

b) \(\sqrt{10}+\sqrt{17}+1>\sqrt{9}+\sqrt{16}+1=8=\sqrt{64}>\sqrt{61}\)

c) \(\left(\sqrt{2016}+\sqrt{2014}\right)^2=4030+\sqrt{2014.2016}\)

\(\left(2\sqrt{2015}^2\right)=4030+\sqrt{2015.2015}\)

C/m được: \(\sqrt{2014.2016}< \sqrt{2015.2015}\)

\(\Rightarrow\left(\sqrt{2016}+\sqrt{2014}\right)^2< \left(2\sqrt{2015}\right)^2\)

\(\Rightarrow\sqrt{2014}+\sqrt{2016}< 2\sqrt{2015}\)

d) \(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=7=8-1=\sqrt{64}-1< \sqrt{65}-1\)

vế phải lớn hơn nha

bn giải cụ thể giúp mk vs đc ko