Ta có : \(B=\dfrac{1}{12}>\dfrac{1}{22};\dfrac{1}{13}>\dfrac{1}{22};....;\dfrac{1}{21}>\dfrac{1}{22}\)

Vậy : \(B=\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{22}>\dfrac{1}{22}+\dfrac{1}{22}+\dfrac{1}{22}+...+\dfrac{1}{22}=\dfrac{11}{22}=\dfrac{1}{2}\)

( Có 11 số hạng \(\dfrac{1}{2}\))

Hay B \(>\dfrac{1}{2}\)

Lời giải:

Ta có:

\(\left\{\begin{matrix} \frac{1}{13}< \frac{1}{12}\\ \frac{1}{14}< \frac{1}{12}\\ \frac{1}{15}< \frac{1}{12}\end{matrix}\right.\Rightarrow \frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{3}{12}=\frac{1}{4}(1)\)

\(\left\{\begin{matrix} \frac{1}{61}< \frac{1}{60}\\ \frac{1}{62}< \frac{1}{60}\\ \frac{1}{63}< \frac{1}{60}\end{matrix}\right.\Rightarrow \frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{3}{60}=\frac{1}{20}(2)\)

Từ \((1);(2)\Rightarrow \frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}\)

Hay \( \frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)

Ta có đpcm.

Đặt A là biểu thức đó

Ta có:

\(\dfrac{1}{13}< \dfrac{1}{12};\dfrac{1}{14}< \dfrac{1}{12};\dfrac{1}{15}< \dfrac{1}{12}\)

\(\Rightarrow\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}\)

Ta cũng có

\(\dfrac{1}{61}< \dfrac{1}{60};\dfrac{1}{62}< \dfrac{1}{60};\dfrac{1}{63}< \dfrac{1}{60}\)

\(\Rightarrow\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{12}.3+\dfrac{1}{60}.3\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

\(\Rightarrow\)dpcm

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{20}-\dfrac{1}{20}\right)\)

\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)

\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)

\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\) (đpcm)

Giải

Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)

\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)

Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)

\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

D< 1 - \(\dfrac{1}{20}\)

D< \(\dfrac{19}{20}\)<1

\(\Rightarrow\)D< 1

Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1

A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)

A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)

\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)

Ta có :

\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :

\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)

\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)

A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1

A<\(\dfrac{49}{200}< \dfrac{1}{2}\)

\(\Rightarrow A< \dfrac{1}{2}\)

có gì sai xin mấy bạn chỉ bảo!!!

bài 2

Ta có: A\(=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)

\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)

\(=\dfrac{1}{9}-\dfrac{1}{15}=\dfrac{2}{45}\)

\(A=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)

\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)

\(=\dfrac{1}{9}-\dfrac{1}{15}\)

\(=\dfrac{2}{45}\)

-Chúc bạn học tốt-

A = \(\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)

= \(\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+\dfrac{1}{13.14}+\dfrac{1}{14.15}\)

= \(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+...+\dfrac{1}{14}-\dfrac{1}{15}\)

= \(\dfrac{1}{9}-\dfrac{1}{15}\)

= \(\dfrac{2}{45}\)

\(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}-\dfrac{1}{3}\)

\(=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)-\dfrac{1}{3}\)

\(=-1+1-\dfrac{1}{3}\)

\(=0-\dfrac{1}{3}\)

\(=\dfrac{-1}{3}\)

------------------------------------------

\(14.\dfrac{3}{2}+\dfrac{6}{5}:\left(-\dfrac{2}{5}\right)\)

\(=14.\dfrac{3}{2}+\dfrac{6}{5}.\dfrac{-5}{2}\)

\(=21+\dfrac{6}{5}.\dfrac{-5}{2}\)

\(=21+\left(-3\right)\)

\(=18\)

------------------------------------------------

\(\sqrt{\dfrac{1}{4}+\dfrac{2}{3}-\left(\dfrac{1}{3}\right)^2}\)

\(=\sqrt{\dfrac{1}{4}+\dfrac{2}{3}-\dfrac{1}{9}}\)

\(=\sqrt{\dfrac{3}{12}+\dfrac{8}{12}-\dfrac{1}{9}}\)

\(=\sqrt{\dfrac{11}{12}-\dfrac{1}{9}}\)

\(=\sqrt{\dfrac{99}{108}-\dfrac{12}{108}}\)

\(=\sqrt{\dfrac{29}{36}}\)

\(=\dfrac{\sqrt{29}}{6}\)

\(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\dfrac{5}{4}+\dfrac{5}{13}-\dfrac{1}{4}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\left(\dfrac{5}{4}-\dfrac{1}{4}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)-\dfrac{1}{3}\)
\(=1+\left(-1\right)-\dfrac{1}{3}=0-\dfrac{1}{3}=-\dfrac{1}{3}\)

\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)

\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)

\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)

Vậy \(x\in\left\{22;23;24;...\right\}\)

\(\dfrac{????????}{????????????}\)

a) \(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)

   \(\dfrac{11}{55}+\dfrac{10}{55}< \dfrac{x}{55}< \dfrac{22}{55}+\dfrac{1}{55}\)

   \(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{23}{55}\)

\(\Rightarrow\) \(x=22\)

b) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}< x\le\dfrac{13}{4}+\dfrac{14}{8}\)

  \(\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6}< x\le\dfrac{26}{8}+\dfrac{14}{8}\)

  \(1< x\le5\)

  \(\Rightarrow\) \(x\in\) {\(2;3;4;5\)}

c) \(\dfrac{1}{3}+\dfrac{5}{12}+\dfrac{-1}{13}< x< \dfrac{7}{5}+\dfrac{2}{10}+\dfrac{1}{2}\)

 Ko biết làm

d) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)

   \(\dfrac{79}{15}+\dfrac{21}{15}+\dfrac{-40}{15}\le x\le\dfrac{40}{12}+\dfrac{45}{12}+\dfrac{23}{12}\)

   \(4\le x\le9\)

   \(\Rightarrow\) \(x\in\) {\(4;5;6;7;8;9\)}