11.

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

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

\(...\)

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{100}\)

\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{49}{100}< \dfrac{50}{100}< \dfrac{1}{2}\) (ĐPCM)

12.

Ta có: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}>0_{\left(1\right)}.\)(Do A là phân số).

Ta lại có:

\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}.\)

\(=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}.\)

\(= \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}.\)

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

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

\(= 1+0+0+0+...+0-\dfrac{1}{100}.\)

\(= 1-\dfrac{1}{100}.\)

\(= \dfrac{99}{100}.\)

\(\Rightarrow A< 1_{\left(2\right)}.\) (do \(\dfrac{99}{100}< 1\)).

Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\) \(\Rightarrow\) \(0< A< 1.\)

\(\Rightarrow A\) không phải là số nguyên.

Vậy ta thu được \(đpcm.\)

14.

Đặt \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>\dfrac{1}{31}+\dfrac{1}{31}+\dfrac{1}{31}+...+\dfrac{1}{31}\)(có 62 số hạng \(\dfrac{1}{31}\))

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>\dfrac{1}{31}\times62\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>2\)

\(Vậy\) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>2\left(đpcm\right)\)

Đáp án:

Giải thích các bước giải:

11

Ta có:

$\frac{1}{3^2}$ $\leq$ $\frac{1}{2.3}$

$\frac{1}{4^2}$ $\leq$ $\frac{1}{3.4}$

$\frac{1}{5^2}$ $\leq$ $\frac{1}{4.5}$

$...$

$\frac{1}{100^2}$ $\leq$ $\frac{1}{99.100}$

$⇒$$\frac{1}{3^2}$ $+$$\frac{1}{4^2}$$+$ $\frac{1}{5^2}$$+...+$ $\frac{1}{100^2}$$\leq$ 

$\frac{1}{2.3}$ $+$ $\frac{1}{3.4}$ $+...+$ $\frac{1}{99.100}$ $=$ $\frac{1}{2}$ $-$ $\frac{1}{100}$ $<$ $\frac{1}{2}$ 

12 

Ta có:

$\frac{1}{2^2}$ $\leq$ $\frac{1}{1.2}$

$\frac{1}{3^2}$ $\leq$ $\frac{1}{2.3}$

$\frac{1}{4^2}$ $\leq$ $\frac{1}{3.4}$

$...$

$\frac{1}{100^2}$ $\leq$ $\frac{1}{99.100}$

$⇒$$\frac{1}{2^2}$ $+$$\frac{1}{3^2}$$+$ $\frac{1}{4^2}$$+...+$ $\frac{1}{100^2}$$\leq$ 

$\frac{1}{1.2}$$+$$\frac{1}{2.3}$$+...+$ $\frac{1}{99.100}$$=$$\frac{1}{1}$ $-$$\frac{1}{100}$$<$ $1$

và $\frac{1}{1}$ $-$$\frac{1}{100}$$>$0

⇒ A không là số nguyên

14

Ta có

$\frac{1}{2}$ $=$ $\frac{1}{2}$

$\frac{1}{3}$$+$$\frac{1}{4}$>$\frac{1}{4}$$+$$\frac{1}{4}$=$\frac{1}{2}$

1/5+...+1/8>4.1/8=1/2

1/9+1/10+...+1/16>8.1/16=1/2

⇒1/2+1/3+1/4+...+1/16>1/2+1/2+1/2+1/2=2

Mà 1/2+1/3+1/4+...+1/63>1/2+1/3+1/4+...+1/16

⇒1/2+1/3+1/4+...+1/63>2