a) \(\Leftrightarrow\left|2x-3\right|=\frac{1}{4}\Leftrightarrow\orbr{\begin{cases}x\ge\frac{3}{2}\mid:2x-3=\frac{1}{4}\Rightarrow2x=\frac{13}{4}\Rightarrow x=\frac{13}{8}\left(TM\right)\\x< \frac{3}{2}\mid:3-2x=\frac{1}{4}\Rightarrow2x=\frac{11}{4}\Rightarrow x=\frac{11}{8}\left(TM\right)\end{cases}.}\)

b) \(\Leftrightarrow\left|x-1\right|=\frac{3}{4}\Leftrightarrow\orbr{\begin{cases}x\ge1\mid:x-1=\frac{3}{4}\Rightarrow x=\frac{7}{4}\left(TM\right)\\x< 1\mid:1-x=\frac{3}{4}=>x=\frac{1}{4}\left(TM\right)\end{cases}}\)

c) \(\frac{3}{5\left(x-\frac{5}{6}\right)}-\frac{1}{2\left(\frac{3}{2}-1\right)}=-\frac{1}{4}\Leftrightarrow\frac{3}{\frac{5\left(6x-5\right)}{6}}-\frac{1}{2\cdot\frac{1}{2}}=-\frac{1}{4}\Leftrightarrow\frac{18}{5\left(6x-5\right)}=-\frac{1}{4}+1\)

\(\Leftrightarrow\frac{18}{5\left(6x-5\right)}=\frac{3}{4}\Leftrightarrow6x-5=\frac{24}{5}\Leftrightarrow6x=\frac{49}{5}\Leftrightarrow x=\frac{49}{30}\)

d) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(\Leftrightarrow\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(\Leftrightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\Leftrightarrow2\cdot\frac{x+1-2}{2\left(x+1\right)}=\frac{2015}{2016}\Leftrightarrow\frac{x-1}{x+1}=\frac{2015}{2016}\)

\(\Leftrightarrow2016x-2016=2015x+2015\Leftrightarrow x=2015+2016=4031\)

Vậy x = 4031.

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{x\left(x+1\right):2}=\frac{2001}{2003}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003}:2=\frac{2001}{4006}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}=\frac{1}{2003}\)

=> x+1 = 2003

=> x = 2003 - 1

=> x = 2002

thank you very much

1: \(=\dfrac{1}{3}\cdot\dfrac{3}{7}\cdot\dfrac{1}{2}=\dfrac{1}{7\cdot2}=\dfrac{1}{14}\)

2: =5/12+1/7

=35/84+12/84=47/84

3: =8(8/9-6/9)

=8*2/9=16/9

4: \(=\dfrac{5}{12}\cdot\dfrac{12}{5}=1\)

5: =16/5+6

=16/5+30/5=46/5

6: =10*1/2-10*1/5

=5-2=3

a)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{10}+x=\frac{3}{5}\)

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

\(\Rightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)

b)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{13}-\frac{2}{15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{2}{3}-\frac{2}{15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{8}{15}+x=\frac{1}{3}\)

\(\Rightarrow x=\frac{1}{3}-\frac{8}{15}=-\frac{1}{5}\)

c)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)

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

\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)

\(\Leftrightarrow\frac{x+1-1}{x+1}=\frac{9}{10}\)

\(\Rightarrow\frac{x}{x+1}=\frac{9}{10}\)

\(\Rightarrow x=9\)

b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{15}\)

a =143

b = 2510

c 28 x 73 = 2044

cái câu c tính nhẩm chết liền

d [ 1700 - 340 +17 ] : 17 = 81

2 . Tìm x 

a 7 . [ x - 3 ] - 6 = 1

   7 . [ x - 3 ]   = 1+6

7 . [ x - 3 ]     = 7

     x - 3 = 7 : 7

x - 3 =1

x=3+1=4

b 

[x+5] .2:11=6

[x+5] x 2  = 6 x 11

[ x + 5 ] x 2 = 66

x+2=66:2

x+2=33

x=33-2=31

Bài 1 

a, \(45+98=143\)

b, \(2605-95=2510\)

c,\(28.73+28\)

\(=28.\left(73+1\right)\)

\(=28.74\)

\(=2072\)

d,\(\left[1700-340+17\right]:17\)

\(=1700:17-340:17+17:17\)

\(=100-20+1\)

\(=81\)

Bài 2 

a, \(7.\left(x-3\right)-6=1\)

\(\Leftrightarrow7.\left(x-3\right)=7\)

\(\Leftrightarrow x-3=1\)

\(\Leftrightarrow x=4\)

b, \(\left(x+2\right).2:11=6\)

\(\Leftrightarrow\left(x+2\right).2=66\)

\(\Leftrightarrow x+2=33\)

\(\Leftrightarrow x=31\)