a: \(2\cdot\sqrt{31}>2\cdot\sqrt{25}=2\cdot5=10\)

=>\(-2\sqrt{31}<-10\)

b: \(2\sqrt3-5-\left(\sqrt5-4\right)=2\sqrt3-\sqrt5-5+4=2\sqrt3-\sqrt5-1\)

\(\left(2\sqrt3\right)^2=12;\left(\sqrt5+1\right)^2=6+2\sqrt5\)

=>\(\left(2\sqrt3\right)^2-\left(\sqrt5+1\right)^2=12-6-2\sqrt5=6-2\sqrt5>0\)

=>\(\left(2\sqrt3-\sqrt5-1\right)\left(2\sqrt3+\sqrt5+1\right)>0\)

mà \(\left(2\sqrt3+\sqrt5+1\right)>0\)

nên \(\left(2\sqrt3-\sqrt5-1\right)>0\)

=>\(2\sqrt3-5-\left(\sqrt5-4\right)>0\)

=>\(2\sqrt3-5>\sqrt5-4\)

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 \(4\sqrt{5}>5\sqrt{3}\)

4 8/3 và 3\(\sqrt2\)

4\(\frac83\) = \(\frac{20}{3}\) = \(\sqrt{\frac{400}{3}}\)

3\(\sqrt2\) = \(\sqrt{9.2}\) = \(\sqrt{18}\) = \(\sqrt{\frac{54}{3}}\)

Vì 400/3 > 54/3 nên \(\sqrt{\frac{400}{3}}\) > \(\sqrt{\frac{54}{3}}\)

Vậy: 4\(\frac83\) > 3\(\sqrt2\)

So sánh:

5\(\sqrt{\left(-10\right)^2}\) và 10\(\sqrt{\left(-5\right)^2}\)

5\(\sqrt{\left(-10\right)^2}\) = \(\sqrt{100.5^2}\) = \(\sqrt{2500}\)

10\(\sqrt{\left(-5\right)^2}\) = \(\sqrt{10^2.25}\) = \(\sqrt{2500}\)

Vậy 5\(\sqrt{\left(-10\right)^2}\) = 10\(\sqrt{\left(-5\right)^2}\)

a) <

b) <

c) >

d) <

      a <

            b <

                           c >

                   d <

Bài 1: 

Để M có nghĩa thì \(\left\{{}\begin{matrix}x+4\ge0\\2-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x\le2\end{matrix}\right.\Leftrightarrow-4\le x\le2\)

Số giá trị nguyên thỏa mãn điều kiện là:

\(\left(2+4\right)+1=7\)

dell bt

a: \(1< \sqrt{2}\)

nên \(2< \sqrt{2}+1\)

b: \(2\sqrt{31}=\sqrt{124}\)

\(10=\sqrt{100}\)

mà 124>100

nên \(2\sqrt{31}>10\)

c: \(-3\sqrt{11}=-\sqrt{99}\)

\(-\sqrt{12}=-\sqrt{12}\)

mà 99>12

nên \(-3\sqrt{11}< -\sqrt{12}\)