\(=1+2+3+...+20+21\)
=> Tổng trên có 21 số
Tổng là : (21+1).21/2=231
\(\sqrt{1}+\sqrt{4}+\sqrt{9}+....+\sqrt{400}+\sqrt{441}\)
= 1+2+3+...+21=231
Tính \(\sqrt{1}-\sqrt{4}+\sqrt{9}-\sqrt{16}+\sqrt{25}-\sqrt{36}+.....-\sqrt{400}\)
A = \(\sqrt{\frac{1}{9}+\frac{1}{16}}\)
B = \(\sqrt{4+36+81}\)
C = \(\sqrt{1^{3^{ }}+}2^3\)
E = \(\left(\sqrt{\frac{1}{9}+\sqrt{\frac{25}{36}-\sqrt{\frac{49}{81}}:}\sqrt{\frac{441}{324}}}\right)\)
F = \(-4.\sqrt{\frac{1}{16}+3.\sqrt{\frac{1}{9}}-}5.\sqrt{0,04}\)
Mg các bạn giúp mik ạ
Cảm ơn ạ
5\(\sqrt{16}\)-4\(\sqrt{9}\)+\(\sqrt{25}\)-0,3\(\sqrt{400}\)
E = \(5\sqrt{16}-4\sqrt{9}+\sqrt{25}-0,3\sqrt{400}\)
\(5\sqrt{16}-4\sqrt{9}+\sqrt{25}-\sqrt{400}\)
\(5\sqrt{16}-4\sqrt{9}+\sqrt{25}-0,3\sqrt{400}\)
\(5\sqrt{16}-4\sqrt{9}+\sqrt{25}-0,3\sqrt{400}\)
Hãy chứng minh:
\(\sqrt{17}+\sqrt{26}+\sqrt{101}>\sqrt{441}\)
Tính: \(\sqrt{1}\) - \(\sqrt{4}\) + \(\sqrt{9}\) - \(\sqrt{16}\) + \(\sqrt{25}\) - \(\sqrt{36}\) +..............- \(\sqrt{400}\)
(Giải nhanh nhé,gấp lắm)