\(1)B=1+3+3^2+...+3^{1999}+3^{2000}\\3B=3+3^2+3^3+...+3^{2000}+3^{2001}\\3B-B=3+3^2+3^3+...+3^{2000}+3^{2001}-(1+3+3^2+...+3^{1999}+3^{2000})\\2B=3^{2001}-1\\\Rightarrow B=\dfrac{3^{2001}-1}{2}\\---\)

\(2)C=1+4+4^2+...+4^{99}+4^{100}\\4C=4+4^2+4^3+...+4^{100}+4^{101}\\4C-C=4+4^2+4^3+...+4^{100}+4^{101}-(1+4+4^2+....+4^{99}+4^{100})\\3C=4^{101}-1\\\Rightarrow C=\dfrac{4^{101}-1}{3}\)

#\(Toru\)

1) \(B=1+3+3^2+...+3^{1999}+3^{2000}\)

\(3B=3\cdot\left(1+3+3^2+...+3^{2000}\right)\)

\(3B=3+3^2+...+3^{2001}\)

\(3B-B=3+3^2+3^3+...+3^{2001}-1-3-3^2-...-3^{2000}\)

\(2B=3^{2001}-1\)

\(B=\dfrac{3^{2001}-1}{2}\)

2) \(C=1+4+4^2+...+4^{100}\)

\(4C=4\cdot\left(1+4+4^2+...+4^{100}\right)\)

\(4C=4+4^2+4^3+...+4^{101}\)

\(4C-C=4+4^2+4^3+...+4^{201}-1-4-4^2-....-4^{100}\)

\(3C=4^{101}-1\)

\(C=\dfrac{4^{101}-1}{3}\)

Mình cho bạn công thức tổng quát để sau này tiện áp dụng nhé:

\(A=1+a^1+a^2+...+a^n\)

\(\Rightarrow A=\dfrac{a^{n+1}-1}{a-1}\)

A. ta  có \(5=\sqrt{25}\)

vì \(\sqrt{25}< \sqrt{29}\)

suy ra \(5< \sqrt{29}\)

k cho mk nha

waaaaaaaakholuon

5=\(\sqrt{25}\)<\(\sqrt{29}\)(do 25<29)

\(3\sqrt{2}\)=\(\sqrt{18}\);\(2\sqrt{3}\)=\(\sqrt{12}\)

Nhu vay \(3\sqrt{2}\)>\(2\sqrt{3}\)

\(\sqrt{35}+\sqrt{99}< \sqrt{36}+\sqrt{100}=6+10=16\)

Vậy \(\sqrt{35}+\sqrt{99}< 16\)

A = \(\dfrac{5^{2020}+1}{5^{2021}+1}\) ⇒ A \(\times\) 10 = 2 \(\times\)5 \(\times\) \(\dfrac{5^{2020}+1}{5^{2021}+1}\) =2\(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\)

10A =2 \(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\) = 2 \(\times\)(1 + \(\dfrac{4}{5^{2021}+1}\) )= 2 + \(\dfrac{8}{5^{2021}+1}\) >2

B = \(\dfrac{10^{2019}+1}{10^{2020}+1}\) ⇒ B \(\times\) 10 = 10 \(\times\) \(\dfrac{10^{2019}+1}{10^{2020}+1}\)= \(\dfrac{10^{2020}+10}{10^{2020}+1}\)

10B = \(\dfrac{10^{2020}+10}{10^{2020}+1}\) = 1 + \(\dfrac{9}{10^{2020}+1}\) < 2

10A > 2 > 10B ⇒ 10A>10B ⇒ A>B

theo ket qua cho thay:9.4594<10

Ta có :

\(\sqrt{3}< \sqrt{4}=2\)

\(\sqrt{8}< \sqrt{9}=3\)

\(\sqrt{24}< \sqrt{25}=5\)

\(\Rightarrow\sqrt{3}+\sqrt{8}+\sqrt{24}< 2+3+5=10\)(đpcm)

Vậy ...

\(\sqrt{3}+\sqrt{8}+\sqrt{24}< \sqrt{4}+\sqrt{9}+\sqrt{25}\)

\(=2+3+5=10\)

Vậy: \(\sqrt{3}+\sqrt{8}+\sqrt{24}< 10\)

\(\sqrt{27}-\sqrt{12}-\sqrt{2016}>\sqrt{25}-\sqrt{16}-\sqrt{2025}\)

\(=5-4-45=-44\)

Vậy \(\sqrt{27}-\sqrt{12}-\sqrt{2016}>-44\)

Có : \(\sqrt{12}< \sqrt{16}=4\)

         \(\sqrt{2016}< \sqrt{2025}\)         => \(\sqrt{12}+\sqrt{2016}< 4+45\)

                                                                 => \(-\sqrt{12}-\sqrt{2016}>-49\)(1)

Lại có : \(\sqrt{27}>\sqrt{25}=5\)(2)

Từ (1),(2) có : \(\sqrt{27}-\sqrt{12}-\sqrt{2016}>5-49\)or \(\sqrt{27}-\sqrt{12}-\sqrt{2016}>-44\)

chơi nhau à

có cc=)))

Giải hộ đi mà =//

ta có : 

A = \(\dfrac{5^{2020}+1}{5^{2020}+1}\)

B = \(\dfrac{5^{2019}+1}{5^{2020}+1}\)

\(\Leftrightarrow\) B < A