Đặt \(A=\frac{4}{19.21}+\frac{4}{21.23}+\frac{12}{23.29}+\frac{4}{29.31}\)

\(\Rightarrow\frac{1}{2}A=\frac{2}{19.21}+\frac{2}{21.23}+\frac{6}{23.29}+\frac{2}{29.31}\)

\(\Rightarrow\frac{1}{2}A=\frac{21-19}{19.21}+\frac{23-21}{21.23}+\frac{29-23}{23.29}+\frac{31-29}{29.31}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{19}-\frac{1}{21}+\frac{1}{21}-\frac{1}{23}+\frac{1}{23}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{19}-\frac{1}{31}=\frac{12}{589}\)

\(\Rightarrow A=\frac{12}{589}:\frac{1}{2}=\frac{24}{589}\)

\(M=\dfrac{6}{10.13}+\dfrac{6}{13.16}+\dfrac{6}{16.19}+\dfrac{6}{19.21}\)

\(\dfrac{1}{2}M=\dfrac{3}{10.13}+\dfrac{3}{13.16}+\dfrac{3}{16.19}+\dfrac{3}{19.21}\)

\(\dfrac{1}{6}M=\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{21}\)

\(\dfrac{1}{6}M=\dfrac{1}{10}-\dfrac{1}{21}\)

\(M=\dfrac{11}{210}:\dfrac{1}{6}=\dfrac{11}{35}\)

\(N=\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{30}\)

\(=\dfrac{1}{20}-\dfrac{1}{30}\)

\(=\dfrac{1}{60}\)

\(\dfrac{M}{N}=\dfrac{11}{35}:\dfrac{1}{60}=\dfrac{132}{7}\)= \(\dfrac{132}{25}\)

\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{19.21}\)

\(=2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{19.21}\right)\)

\(=2\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)

\(=2\left(1-\dfrac{1}{21}\right)=2.\dfrac{20}{21}=\dfrac{40}{21}\)

a) Ta có : \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\\ x+2=0\Rightarrow x=-2\)

Lập bảng xét dấu:

TH : Xét x < -2

Ta có : - ( x+ 2) - (x - \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)

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

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

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

=> x = \(-1\dfrac{1}{8}\) ( loại )

TH 2: \(-2\le x< \dfrac{1}{2}\)

Ta có : x + 2 + ( -x + \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)

=> \(2,5=\dfrac{3}{4}\) ( loại )

TH3 : \(x\ge\dfrac{1}{2}\)

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

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

x = -0,375( loại )

vậy ....

b) \(\left(\dfrac{2}{3}-2x\right).1\dfrac{1}{2}=\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{3}-2x=-\dfrac{3}{4}\\ \Rightarrow2x=1\dfrac{5}{12}\\ \Rightarrow x=\dfrac{17}{24}\)

c) \(\left|x-1\right|+2.\left(x+4\right)=10\\ \Rightarrow\left|x-1\right|=10-2x-8\\ \Rightarrow\left|x-1\right|=2-2x\)

TH1 : \(x-1\ge0\) \(\Rightarrow x\ge1\)

\(\Rightarrow x-1=2-2x\\ \Rightarrow3x=3\\ \Rightarrow x=1\left(TM\right)\)

TH2 : \(x-1< 0\Rightarrow x< 1\)

=> \(x-1=-2+2x\\ \Rightarrow-x=-1\Rightarrow x=1\)(loại)

Vậy x = 1

b. \(\left(\dfrac{2}{3}-2x\right)\cdot1\dfrac{1}{2}=\dfrac{3}{4}\Rightarrow\dfrac{2}{3}-2x=\dfrac{3}{4}:\dfrac{3}{2}=\dfrac{1}{2}\Rightarrow-2x=\dfrac{1}{2}-\dfrac{2}{3}=-\dfrac{1}{6}\Rightarrow x=-\dfrac{1}{6}:\left(-2\right)=\dfrac{1}{12}\)

Vậy \(x=\dfrac{1}{12}\)

c. \(\left|x-1\right|+2\left(x+4\right)=10\Rightarrow\left|x-1\right|+2x+8=10\Rightarrow\left|x-1\right|+2x=10-8=2\)

\(\Rightarrow\left[{}\begin{matrix}x-1+2x=2;x-1\ge0\\-\left(x-1\right)+2x=2;x-1< 0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+2x=2+1+3;x\ge1\\-x+1+2x=2;x< 1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=3;x\ge1\\-x+2x=2-1=1;x< 1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3:3=1;x\ge1\\x=1;x< 1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x\in\varnothing\end{matrix}\right.\)

Vậy x = 1

d. \(\dfrac{11}{12}+\dfrac{11}{12\cdot23}+...+\dfrac{11}{89\cdot100}+x=1\dfrac{2}{3}\)

\(\Rightarrow\dfrac{11}{12}+\dfrac{11}{276}+...+\dfrac{11}{8900}+x=\dfrac{5}{3}\)

\(\Rightarrow\dfrac{22}{23}+...+\dfrac{11}{8900}+x=\dfrac{5}{3}\)

\(\Rightarrow\dfrac{99}{100}+x=\dfrac{5}{3}\Rightarrow x=\dfrac{5}{3}-\dfrac{99}{100}=\dfrac{203}{300}\)

Vậy \(x=\dfrac{203}{300}\)

e. \(\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)-x+4\dfrac{221}{231}=2\dfrac{1}{3}\)

\(\Rightarrow\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)-x=\dfrac{7}{3}-4\dfrac{221}{231}\)

\(\Rightarrow x=\dfrac{\dfrac{7}{3}-\dfrac{1145}{231}}{\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}}=\dfrac{-\dfrac{202}{77}}{\dfrac{2}{143}+\dfrac{2}{195}+...+\dfrac{2}{399}}=\dfrac{-\dfrac{202}{77}}{\dfrac{10}{231}}=\dfrac{-202}{77}\cdot\dfrac{231}{10}=\dfrac{-303}{5}\)

Vậy \(x=-\dfrac{303}{5}\)

\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)

\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)

=0

    Mới thế đã hai năm trôi qua,câu trả lời từ mọi người vẫn KO XUẤT HIỆN.

    Ko biết sau này câu trả lời có xuất hiện hay ko...

\(S=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{29\cdot31}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{29}-\dfrac{1}{31}\\ =\dfrac{1}{1}-\dfrac{1}{31}\\ =\dfrac{30}{31}\)

mà \(\dfrac{30}{31}>\dfrac{2014}{2015}\Rightarrow S>P\)

So sánh vs j nhỉ .-.?

`S=1-1/3+1/3-1/5+...+1/29-1/31`

`S=1-1/31=30/31`

S=2.(1/1-1/3+1/3-1/5+1/5-1/7+...+1/27-1/29+1/29-1/31)

S=2.(1-1/31)

S=2.30/31

S=60/31

P=2014/2015

=>S>P hay 60/31 > 2014 / 2015

Mình đang cần gấp!☹

Đ V T H

a: =9/12-1/12-2/12+2=1/2+2=5/2

b: =(2/5+4/5+3/10)-5/6=6/5+3/10-5/6=15/10-5/6=3/2-5/6=9/6-5/6=4/6=2/3

c: \(=\dfrac{12}{25}\cdot\dfrac{32}{9}\cdot\dfrac{15}{16}=\dfrac{12}{9}\cdot\dfrac{32}{16}\cdot\dfrac{15}{25}=\dfrac{4}{3}\cdot\dfrac{3}{5}\cdot2=\dfrac{8}{5}\)

a) \(1+\dfrac{4}{9}=\dfrac{9}{9}+\dfrac{4}{9}=\dfrac{9+4}{9}=\dfrac{13}{9}\)

b) \(5+\dfrac{1}{2}=\dfrac{10}{2}+\dfrac{1}{2}=\dfrac{10+1}{2}=\dfrac{11}{2}\)

c) \(3-\dfrac{5}{6}=\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{18-5}{6}=\dfrac{13}{6}\)

d) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)