2/3.5+2/5.7+2/7.9+...+2/(2x+1)(2x+3)=2.15/93

1/3-1/5+1/5-1/7+...+1/2x+1-1/2x+3=10/31

1/3-1/2x+3=10/31

1/(2x+3)=1/93

2x+3=93

2x=90

x=45

Đặt A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{5}{31}\)

  2A   = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)

  2A   = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{\left(2x+1\right)}-\frac{1}{2x+3}=\frac{10}{31}\)

  2A   = \(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

  Ta có :  \(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

                           \(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)

                          \(\frac{1}{2x+3}=\frac{1}{93}\)

=> 2x + 3 = 93

     2x       = 90

       x       = 45

1.a) \(\left(31\frac{6}{13}+5\frac{9}{41}\right)-36\frac{6}{13}=\left(31+\frac{6}{13}+5+\frac{9}{41}\right)-\left(36+\frac{6}{13}\right)\)

\(=\left(36+\frac{6}{13}-\frac{9}{41}\right)-\left(36+\frac{6}{13}\right)=\left(36+\frac{6}{13}\right)-\left(36+\frac{6}{13}\right)-\frac{9}{41}=-\frac{9}{41}\)

b) \(\frac{5}{3}+\left(-\frac{2}{7}\right)-\left(-1,2\right)-\left|1.4-0,2\right|\)

\(=\frac{5}{3}-\frac{2}{7}+1,2-1,2=\frac{29}{21}\)

c) \(0,25+\frac{3}{5}-\left(\frac{1}{8}-\frac{2}{5}+1\frac{1}{4}\right)+\left|\frac{3}{5}\right|\)

\(=\frac{1}{4}+\frac{3}{5}-\frac{1}{8}+\frac{2}{5}-1-\frac{1}{4}+\frac{3}{5}\)

\(=\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{3}{5}+\frac{2}{5}-1\right)+\frac{3}{5}-\frac{1}{8}=\frac{19}{40}\)

2) \(-\frac{3}{5}-x=0,75\)

=> \(-\frac{3}{5}-x=\frac{3}{4}\)

=> \(x=-\frac{3}{5}-\frac{3}{4}=\frac{-27}{20}\)

b) \(x+\frac{1}{3}=\frac{2}{5}-\left(-\frac{1}{3}\right)\)

=> \(x+\frac{1}{3}=\frac{2}{5}+\frac{1}{3}\)

=> \(x=\frac{2}{5}\)

c) |2x - 4| + 1 = 5

=> |2x - 4| = 4

<=> \(\orbr{\begin{cases}2x-4=4\\2x-4=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)

Giúp mình với nha cả nhả :<

Cả nhà làm vài ý thui cx được ạ :<

1/2(2/3.5+2/5.7+2/7.9+...+2/(2x+1)(2x+3))=15/93

1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/2x+1-1/2x+3)=15/93

1/2(1/3-1/2x+3)=15/93

=>1/3-1/2x+3=10/31

=>1/2x+3=1/93

=>2x+3=93

2x=93-3=90

=>x=45

Đặt \(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(\Rightarrow2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{93}\)

\(\Rightarrow2x+3=93\)

\(2x=90\)

\(x=45\)

Vậy \(x=45\).

\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

=> \(\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{5}{31}\)

=> \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{5}{31}\)

=> \(\frac{1}{3}-\frac{1}{2x+3}=\frac{5}{31}:\frac{1}{2}=\frac{5}{31}\cdot2=\frac{10}{31}\)

=> \(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}=\frac{1}{93}\)

=> 2x +3  = 93

=> 2x = 90 => x = 45

2\3x-780\11:[13\2.(1\3.5+1\5.7+1\7.9+1\9.11]=-5

2\3x-780\11:[13\2.(1\3-1\5+1\5-1\7+....+1\9-1\11)]=-5

2\3x-780\11:[13\2.(1\3-1\11)]=-5

2\3x-780\11:[13\2.8\33]=-5

2\3x-780\11:52\33=-5

2\3x-525\13=-5

2\3x=-5+525\13

2\3x=460\13

x=460\13:2\3

x=690\13

bài 1

[(x+2)/10¹⁰]+ [(x+2)/11¹¹]= [(x+2)/12¹²]+[(x+2)/13¹³]

=>[(x+2)/10¹⁰]+[(x+2)/11¹¹] - [(x+2)/12¹²]-[(x+2)/13¹³] = 0

=>(x+2).[(1/10¹⁰)+(1/11¹¹)-(1/12¹²)-(1/13¹³)=0

Vì [(1/10¹⁰)+(1/11¹¹)-(1/12¹²)-(1/13¹³)] khác 0

=>x+2=0

=>x=-2

Bài 1: x=-2

Bài 2:x=17

Bài 3:x=2014

y=2010

Bài 1 : -2

Bài 2 : 15

Bải 3 : x =2014 ; y = 2010

mn giúp

Này bạn làm sao để ra dấu phân số vậy

Bài 1:

a) (2x-3). (x+1) < 0

=>2x-3 và x+1 ngược dấu

Mà 2x-3<x+1 với mọi x

\(\Rightarrow\begin{cases}2x-3< 0\\x+1>0\end{cases}\)

\(\Rightarrow\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)\(\Rightarrow-1< x< \frac{3}{2}\)

b)\(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)

\(\Rightarrow x-\frac{1}{2}\) và x+3 cùng dấu

Xét \(\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\)\(\Rightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\)

Xét \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)\(\Rightarrow\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)

=>....

Bài 2:

\(S=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{999.1001}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{1001}\right)\)

\(=\frac{1}{2}\cdot\frac{998}{3003}\)

\(=\frac{499}{3003}\)