a, \(7+\sqrt{5}\) ta co \(\sqrt{5}>\sqrt{4}\)(1)

\(\sqrt{48}+2\) \(\sqrt{48}<\sqrt{49}\)(2)

\(7+\sqrt{4}=7+2=9\)(3)

\(\sqrt{49}+2=7+2=9\)(4)

tu (1);(2);(3);(3) = > lam not di

b,\(1-\sqrt{50}\) cung so sanh \(\sqrt{50}voi\sqrt{49}\) tu lam not nha

k dung minh nha

Ta có: \(1=\sqrt{1}< \sqrt{50}\Rightarrow1-\sqrt{50}< 0\)

\(\Rightarrow\sqrt{\left(1-\sqrt{50}\right)^2}=\sqrt{50}-1>\sqrt{49}-1=7-1=6\)

Vậy \(\sqrt{\left(1-\sqrt{50}\right)^2}>6\)

ko nhìn thấy bợn ưi

1. Ta có: \(\sqrt{23}+\sqrt{15}< \sqrt{25}+\sqrt{16}=5+4=9\)

mà \(\sqrt{83}>\sqrt{81}=9\)

\(\Rightarrow\sqrt{23}+\sqrt{15}< \sqrt{83}\)

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