:)) ko bt làm :))

                                                                                    kí tên

                                                                                   cái nịt

reeeeeeeee

Ta có :

\(C=\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{1000}}\)

\(\Rightarrow4C=1+\frac{1}{4}+.....+\frac{1}{4^{1999}}\)

\(\Rightarrow4C-C=\left(1+\frac{1}{4}+.....+\frac{1}{4^{1999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{1000}}\right)\)

\(\Rightarrow3C=1-\frac{1}{4^{1000}}\)

\(\Rightarrow C=\frac{1}{3}-\frac{1}{3.4^{1000}}< \frac{1}{3}\)

=> C < 1 / 3

Ta có:

\(C=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{1000}}\)

\(\Rightarrow4C=1+\frac{1}{4}+...+\frac{1}{4^{999}}\)

\(\Rightarrow4C-C=\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}+\frac{1}{4^{1000}}\right)\)

\(\Rightarrow3C=1-\frac{1}{4^{1000}}\)

\(\Rightarrow C=\left(1-\frac{1}{4^{1000}}\right).\frac{1}{3}\)

\(\Rightarrow C=\frac{1}{3}-\frac{1}{4^{1000}.3}\)

Mà \(\frac{1}{3}>\frac{1}{3}-\frac{1}{4^{1000}.3}\)

\(\Rightarrow C< \frac{1}{3}\)

Vậy \(C< \frac{1}{3}\)

Ta có \(\frac{2+3}{4+6}=\frac{1}{2}\)

\(\frac{2-3}{4-6}=\frac{-1}{-2}=\frac{1}{2}\)

=> \(\frac{1}{2}=\frac{1}{2}hay\frac{2+3}{4+6}=\frac{2-3}{4-6}\)

1/100 hả e hay là 1/10

Dạ 1/100

\(A=\frac{1}{\sqrt{2.1}\left(\sqrt{2}+\sqrt{1}\right)}+\frac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+\frac{1}{\sqrt{3.4}\left(\sqrt{4}+\sqrt{3}\right)}+...+\frac{1}{\sqrt{999.1000}\left(\sqrt{1000}+\sqrt{999}\right)}\)

\(A=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2.1}\left(2-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2.3}\left(3-2\right)}+\frac{\sqrt{4}-\sqrt{3}}{\sqrt{3.4}\left(4-3\right)}+...+\frac{\sqrt{1000}-\sqrt{999}}{\sqrt{999.1000}\left(1000-999\right)}\)

\(A=\frac{\sqrt{2}}{\sqrt{2.1}}-\frac{\sqrt{1}}{\sqrt{2.1}}+\frac{\sqrt{3}}{\sqrt{2.3}}-\frac{\sqrt{2}}{\sqrt{2.3}}+\frac{\sqrt{4}}{\sqrt{3.4}}-\frac{\sqrt{3}}{\sqrt{3.4}}+...+\frac{\sqrt{1000}}{\sqrt{999.1000}}-\frac{\sqrt{999}}{\sqrt{1000.999}}\)

\(A=\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{999}}-\frac{1}{\sqrt{1000}}\)

\(A=\frac{1}{1}-\frac{1}{\sqrt{1000}}=\frac{\sqrt{1000}-1}{\sqrt{1000}}=\frac{10\sqrt{10}-1}{10\sqrt{10}}\)

1)

A = \(\frac{3}{8}+\frac{4}{9}+\frac{1}{3}\)

A = \(\frac{27}{72}+\frac{32}{72}+\frac{24}{72}\)
A = \(\frac{83}{72}\)

Vì \(\frac{83}{72}>1\)nên A > 1

B = \(\frac{4}{15}+\frac{4}{13}+\frac{1}{3}\)

B = \(\frac{52}{195}+\frac{60}{195}+\frac{65}{195}\)

B = \(\frac{177}{195}\)

Vì \(\frac{177}{195}< 1\)nên B < 1

a, Ta có : 3/8 > 3/9 = 1/3

                4/9 > 3/9 = 1/3

=> A > 1/3 + 1/3 + 1/3 = 1

b, Ta có : 4/15 < 5/15 = 1/3

                4/13 < 4/12 = 1/3

=> B < 1/3 + 1/3 + 1/3 = 1

Tk mk nha