\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2\cdot2}+\dfrac{1}{2\cdot2}-\dfrac{1}{2\cdot2\cdot2}+\dfrac{1}{2\cdot2\cdot2}-\dfrac{1}{2\cdot2\cdot2\cdot2}+.....+\dfrac{1}{2^{10}}\)

\(A=1-\dfrac{1}{2^{10}}\)

\(A+\dfrac{1}{2^{10}}=1-\dfrac{1}{2^{10}}+\dfrac{1}{2^{10}}=1\left(dpcm\right)\)

a)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)

b)

 \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)

c)

\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)

d) tương tự câu 1

Đặt \(A=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{60^2}\)

\(A< \dfrac{1}{3^2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{59.60}\)

\(A< \dfrac{1}{3^2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\)

\(A< \dfrac{1}{3^2}+\dfrac{1}{3}-\dfrac{1}{60}\)

\(A< \dfrac{4}{9}-\dfrac{1}{60}< \dfrac{4}{9}\) (đpcm)

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

Ta có 

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...............

\(\dfrac{1}{8^2}< \dfrac{1}{7.8}\)

=> B < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{7.8}\)

B < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

B < \(1-\dfrac{1}{8}< 1\) (Do \(\dfrac{1}{8}>0\))

Vậy.....

Ta có :

 \(\dfrac{1}{2}>\dfrac{1}{5}\)

\(\dfrac{1}{3}>\dfrac{1}{5}\)

\(\dfrac{1}{4}>\dfrac{1}{5}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{4}{5}\)

\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\)

\(=\dfrac{7}{6}+\dfrac{1}{4}+\dfrac{1}{5}\)

\(=\dfrac{17}{12}+\dfrac{1}{5}\)

\(=\dfrac{97}{60}\)

\(\dfrac{4}{5}=\dfrac{4.12}{5.12}=\dfrac{48}{60}\)

Mà \(\dfrac{97}{60}>\dfrac{48}{60}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{4}{5}\left(đpcm\right)\).

<3 XD

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

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

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

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

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