Câu hỏi của gam vu - Toán lớp 6 | Học trực tuyến

Có:

\(A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{81}+\dfrac{1}{100}\)

\(A=\dfrac{1}{4}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}+\dfrac{1}{10^2}\)

Mà: \(\dfrac{1}{3^2}>\dfrac{1}{3.4}\)

\(\dfrac{1}{4^2}>\dfrac{1}{4.5}\)

...

\(\dfrac{1}{9^2}>\dfrac{1}{9.10}\)

\(\dfrac{1}{10^2}>\dfrac{1}{10.11}\)

\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)

\(A>\dfrac{1}{4}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)

\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3}-0-0-...-0-\dfrac{1}{11}\)

\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{11}\)

\(\Rightarrow A>\dfrac{65}{132}\)

Chúc bạn học tốt!

Ta có : \(\dfrac{1}{4}\)= \(\dfrac{1}{2.2}\)> \(\dfrac{1}{2.3}\)

\(\dfrac{1}{9}\)= \(\dfrac{1}{3.3}\)> \(\dfrac{1}{3.4}\)

\(\dfrac{1}{16}\)=\(\dfrac{1}{4.4}\)> \(\dfrac{1}{4.5}\)

.......

\(\dfrac{1}{9801}\)= \(\dfrac{1}{99.99}\)> \(\dfrac{1}{99.100}\)

\(\dfrac{1}{10000}\)= \(\dfrac{1}{100.100}\)> \(\dfrac{1}{100.101}\)

\(\Rightarrow\) \(\dfrac{1}{4}\)+ \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+ ..... + \(\dfrac{1}{9801}\)+ \(\dfrac{1}{10000}\)> \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{99.100}\)+\(\dfrac{1}{100.101}\)

= \(\dfrac{3-2}{2.3}\)+ \(\dfrac{4-3}{3.4}\)+ \(\dfrac{5-4}{4.5}\) +...+ \(\dfrac{100-99}{99.100}\)+ \(\dfrac{101-100}{100.101}\)

= \(\dfrac{3}{2.3}\)- \(\dfrac{2}{2.3}\) + \(\dfrac{4}{3.4}\)-\(\dfrac{3}{3.4}\)+ \(\dfrac{5}{4.5}\)-\(\dfrac{4}{4.5}\)+...+ \(\dfrac{100}{99.100}\)- \(\dfrac{99}{99.100}\)+ \(\dfrac{101}{100.101}\)-\(\dfrac{100}{100.101}\)

= \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+....+ \(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)

= \(\dfrac{1}{2}\)- \(\dfrac{1}{101}\) ; Mà \(\dfrac{1}{2}\)- \(\dfrac{1}{101}\)= \(\dfrac{99}{202}\)< \(\dfrac{1}{2}\)

\(\Rightarrow\) \(\dfrac{1}{2}\)< \(\dfrac{1}{4}\)+ \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+...+ \(\dfrac{1}{9801}\)+ \(\dfrac{1}{10000}\) (1)

b\()\)

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4

Tương tự như vậy với câu a\()\)

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 1/2

a) Gọi ƯCLN(12n+1,30n+2) là d

12n+1⋮d  ⇒ 60n+5⋮d 

30n+2⋮d  ⇒ 60n+4⋮d 

(60n+5)-(60n+4)⋮d 

1⋮d 

Vậy \(\dfrac{12n+1}{30n+2}\) là ps tối giản

b) Đặt A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\)

Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A< 1-\dfrac{1}{100}\)

\(A< 1-\dfrac{1}{100}< 1\left(đpcm\right)\)

1: 

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

...

\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+..+\dfrac{1}{7\cdot8}\)

=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{7}{8}< 1\)

A= \(\dfrac{1}{3}-\dfrac{2}{3^2}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

3A= 1 - \(\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+.....+\dfrac{99}{3^{98}}\) - \(\dfrac{100}{3^{99}}\)

A + 3A = 1- \(\dfrac{1}{3}+\dfrac{1}{3^2}\) - \(\dfrac{1}{3^3}+....+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)

=> 4A < 1 - \(\dfrac{1}{3}+\dfrac{1}{3^2}\) \(\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

Đặt : B = 1 - \(\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+....+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

3B = 3 - 1 + \(\dfrac{1}{3}\) - \(\dfrac{1}{3^2}+.....+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}\)

B + 3B = 3 - \(\dfrac{1}{3^{99}}\)

4B = 3 - \(\dfrac{1}{3^{99}}\) < 3 => B < \(\dfrac{3}{4}\)

=> 4A < \(\dfrac{3}{4}\) => A < \(\dfrac{3}{16}\) ĐPCM