\(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) +
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6 tháng 4 2018

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{x\left(x+1\right)}=\dfrac{44}{45}\)

=> \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}+......+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{44}{45}\)

=> \(1-\dfrac{1}{x+1}=\dfrac{44}{45}\)

=> \(\dfrac{x+1}{x+1}-\dfrac{1}{x+1}=\dfrac{44}{45}\)

=> \(\dfrac{x+1-1}{x+1}=\dfrac{44}{45}\)

=> \(\dfrac{x}{x+1}=\dfrac{44}{45}\)

=> 44(x+1)=45x

=> 44x+44=45x

=> 44x-45x=-44

=> -1x=-44

=> x=44

vậy x=44

16 tháng 4 2017

Đặt \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2007\cdot2008}\)

Ta có:

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2008^2}< \)\(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2007\cdot2008}\left(1\right)\)

Lại có: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2007\cdot2008}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)

\(=1-\dfrac{1}{2008}< 1\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\) ta có \(A< B< 1\Rightarrow A< 1\)

16 tháng 4 2017

A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{2008^2}\)

A<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2007.2008}\)

A<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)

A<\(1-\dfrac{1}{2008}\)

A<\(\dfrac{2007}{2008}< 1\)

=> A<1

Vậy A<1

19 tháng 7 2017

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

\(\Leftrightarrow D=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)

\(\Leftrightarrow D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

\(\Leftrightarrow D< \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{10-9}{9.10}\)

\(\Leftrightarrow D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Leftrightarrow D< 1-\dfrac{1}{10}\)

\(\Leftrightarrow D< \dfrac{9}{10}< \dfrac{10}{10}=1\)

\(\Leftrightarrow D< 1\left(đpcm\right)\)

19 tháng 7 2017

Các phần còn lại tương tự như a).

26 tháng 5 2017

a, \(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}\)

\(=\dfrac{32}{99}\)

Vậy \(M=\dfrac{32}{99}\)

b, Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2012^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2011.2012}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\)

\(=1-\dfrac{1}{2012}< 1\) (1)

Do mỗi phân số đều lớn hơn 0 nên \(A>0\) (2)

Từ (1), (2) \(\Rightarrow0< A< 1\)

\(\Rightarrow A\notin N\left(đpcm\right)\)

Vậy...

a, \(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{2}{97}-\dfrac{2}{99}\\ =\dfrac{1}{3}-\dfrac{2}{99}=\dfrac{31}{99}\)

3 tháng 5 2017

giúp mk với

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

30 tháng 4 2017

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

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

11 tháng 4 2017

bài 1:

a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)

\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)

\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)

\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)

\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)

\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)

c) Cho mình hỏi x ở đâu vậy ???

d)\(\left(x-5\right):\dfrac{1}{3}=\dfrac{2}{5}\) e)\(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)

\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)

\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)

\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)

\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)

f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)

\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)

\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)

\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)

\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)

\(x=\dfrac{-5}{7}\)

bài 2:

Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)

\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)

\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)

Vậy \(a=-15;b=81\)

9 tháng 5 2018

bạn chép gì vậy????hay là não bạn có vấn đề?

23 tháng 8 2017

\(a.\)

\(\dfrac{1}{3}\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\)

\(\Rightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)

\(\Rightarrow\dfrac{1}{6}-2+\dfrac{3}{5}=-5x+x\)

\(\Rightarrow-4x=-\dfrac{37}{30}\)

\(\Rightarrow4x=\dfrac{37}{30}\)

\(\Rightarrow x=\dfrac{37}{120}\)

\(b.\)

\(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{3}{2}x-\dfrac{3}{2}=x-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{3}{2}-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{2}\)

\(c.\)

\(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\)

\(\Rightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\)

\(\Rightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)

\(\Rightarrow x=\dfrac{19}{21}\)

\(d.\)

\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\)

\(\Rightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)

\(\Rightarrow\dfrac{3}{2}x-\dfrac{39}{10}=x+1\)

\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{39}{10}+1\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{49}{10}\)

\(\Rightarrow x=\dfrac{49}{5}\)

\(e.\)

\(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\)

\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\)

\(\Rightarrow\left[{}\begin{matrix}4x+3=\dfrac{3}{5}\\4x+3=-\dfrac{3}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{12}{5}\\4x=-\dfrac{18}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{9}{10}\end{matrix}\right.\)

23 tháng 8 2017

a) \(\dfrac{1}{3}.\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\Leftrightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)

\(\Leftrightarrow5x-x=-\dfrac{3}{5}-\dfrac{1}{6}+2\Leftrightarrow4x=\dfrac{37}{30}\Leftrightarrow x=\dfrac{\dfrac{37}{30}}{4}=\dfrac{37}{120}\)

vậy \(x=\dfrac{37}{120}\)

b) \(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\Leftrightarrow\dfrac{3}{2}x-x=\dfrac{-3}{4}+1\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2}x=\dfrac{-3}{4}+\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.2=\dfrac{6}{4}=\dfrac{3}{2}\) vậy \(x=\dfrac{3}{2}\)

c) \(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\Leftrightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\Leftrightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)

\(\Leftrightarrow x=\dfrac{\dfrac{19}{12}}{\dfrac{7}{4}}=\dfrac{19}{21}\) vậy \(x=\dfrac{19}{21}\)

d) \(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\Leftrightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)

\(\Leftrightarrow\dfrac{3}{2}x-x=1+\dfrac{5}{2}+\dfrac{7}{5}\Leftrightarrow\dfrac{1}{2}x=\dfrac{49}{10}\Leftrightarrow x=\dfrac{49}{10}.2=\dfrac{49}{5}\)

vậy \(x=\dfrac{49}{5}\)

e) \(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\Leftrightarrow\left|4x+3\right|=\dfrac{\dfrac{6}{15}}{\dfrac{2}{3}}=\dfrac{3}{5}\)

th1 : \(4x+3\ge0\Leftrightarrow4x\ge-3\Leftrightarrow x\ge\dfrac{-3}{4}\)

\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow4x+3=\dfrac{3}{5}\Leftrightarrow4x=\dfrac{3}{5}-3=\dfrac{-12}{5}\)

\(\Leftrightarrow x=\dfrac{\dfrac{-12}{5}}{4}=\dfrac{-3}{5}\left(tmđk\right)\)

th2: \(4x+3< 0\Leftrightarrow4x< -3\Leftrightarrow x< \dfrac{-3}{4}\)

\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow-\left(4x+3\right)=\dfrac{3}{5}\Leftrightarrow-4x-3=\dfrac{3}{5}\)

\(\Leftrightarrow4x=-3-\dfrac{3}{5}=\dfrac{-18}{5}\Leftrightarrow x=\dfrac{\dfrac{-18}{5}}{4}=\dfrac{-9}{10}\left(tmđk\right)\)

vậy \(x=\dfrac{-3}{5};x=\dfrac{-9}{10}\)