a) Ta có: \(2=\sqrt{4}\)

Vì \(4>3\Rightarrow\sqrt{4}>\sqrt{3}\Rightarrow2>\sqrt{3}\Rightarrow1>\sqrt{3}-1\)

b) \(\left\{{}\begin{matrix}2\sqrt{31}=\sqrt{4.31}=\sqrt{124}\\10=\sqrt{100}\end{matrix}\right.\)

Vì \(124>100\Rightarrow\sqrt{124}>\sqrt{100}\Rightarrow2\sqrt{31}>10\)

c) Vì \(15< 16\Rightarrow\sqrt{15}< \sqrt{16}\Rightarrow\sqrt{15}-1< \sqrt{16}-1\)

\(\Rightarrow\sqrt{15}-1< 4-1\Rightarrow\sqrt{15}-1< 3\)

Lại có: \(10>9\Rightarrow\sqrt{10}>\sqrt{9}\Rightarrow\sqrt{10}>3\)

\(\Rightarrow\sqrt{10}>\sqrt{15}-1\)

a) 1,2+3.1,3=5,1

b) 0,2+2.0,5=1,2

a) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\\\Rightarrow2\sqrt{31}>10\)

Bài 1: 

a) \(\sqrt{1.44}+3\sqrt{1.69}=1.2+3\cdot1.3=1.2+3.9=5.1\)

b) \(\sqrt{0.04}+2\cdot\sqrt{0.25}=0.2+2\cdot0.5=1.2\)

1)  \(A^2=2+2.\frac{\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}}{2}\)

              \(2+\sqrt{64-15}=2+\sqrt{49}=2+7=9\) mà A>0

=> A=3

2) \(A=\sqrt{4-\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right).\)

 \(A=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)

​​\(A=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)

\(A^2=\left(4+\sqrt{15}\right)\left(16-4\sqrt{15}\right)\)

       \(=4\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=4\)

Mà A >0 

=> A=2

Mà 4>3

=> \(\sqrt{4}=2>\sqrt{3}\)

=> \(A>\sqrt{3}\)