a) $\frac{1}{{10}} + \frac{3}{{10}} = \frac{{1 + 3}}{{10}} = \frac{4}{{10}} = \frac{2}{5}$

b) $\frac{5}{{12}} + \frac{1}{{12}} = \frac{{5 + 1}}{{12}} = \frac{6}{{12}} = \frac{1}{2}$

c) $\frac{3}{2} + \frac{1}{2} = \frac{{3 + 1}}{2} = \frac{4}{2} = 2$

a/

S = 1-2+3-4+5-6+...+2001-2002+2003

   = [-1] +[-1] +...+[-1] +2003

      ------------------------

       1001 số -1

= -1001 +2003 = 1002

b/

A = \(6.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)=6.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)=6.\left(\frac{1}{3}-\frac{1}{2015}\right)=\frac{6.2012}{6045}=\frac{4024}{2015}\)

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)

nên x + 1 = 0 => x = -1

Vậy x = -1

b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(1+\frac{x+4}{2000}+1+\frac{x+3}{2001}=1+\frac{x+2}{2002}+1+\frac{x+1}{2003}\)

\(\frac{2004+x}{2000}+\frac{2004+x}{2001}=\frac{2004+x}{2002}+\frac{2004+x}{2003}\)

\(\frac{2004+x}{2000}+\frac{2004+x}{2001}-\frac{2004+x}{2002}-\frac{2004+x}{2003}=0\)

\(\left(2004+x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

Mà \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\)

nên 2004 + x = 0 => x = -2004

Vậy x = -2004

=))

\(A=\frac{12}{3.5}+\frac{12}{5.7}+...+\frac{12}{2013.2015}\)

\(2A=\frac{24}{3.5}+\frac{24}{5.7}+...+\frac{24}{2013.2015}\)

\(2A=\frac{24}{3}-\frac{24}{5}+\frac{24}{5}-\frac{24}{7}+...+\frac{24}{2013}-\frac{24}{2015}\)

\(2A=8-\frac{24}{2015}\)

\(2A=\frac{8}{1}-\frac{24}{2015}\)

\(2A=\frac{16120}{2015}-\frac{24}{2015}\)

\(2A=\frac{16096}{2015}\)

\(=>A=\frac{16096}{2015}:2\)

\(=>A=\frac{16096}{4030}\)

a) $\frac{2}{3} - \frac{1}{3} = \frac{{2 - 1}}{3} = \frac{1}{3}$                 

b) $\frac{7}{{12}} - \frac{5}{{12}} = \frac{{7 - 5}}{{12}} = \frac{2}{{12}} = \frac{1}{6}$  

c) $\frac{{17}}{{21}} - \frac{{10}}{{21}} = \frac{{17 - 10}}{{21}} = \frac{7}{{21}} = \frac{1}{3}$

Tại sao lại =0??

a) \(\frac{1}{12}+\frac{3}{15}+\frac{11}{12}+\frac{1}{71}-\frac{12}{10}=\left(\frac{1}{12}+\frac{11}{12}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\frac{1}{71}\)

\(=\frac{12}{12}+0+\frac{1}{71}=1+\frac{1}{71}=1\frac{1}{71}=\frac{72}{71}\)

b) \(\frac{2}{3}-4\left(\frac{1}{2}+\frac{3}{4}\right)=\frac{2}{3}-4.\frac{5}{4}=\frac{2}{3}-5=\frac{2}{3}-\frac{15}{3}=-\frac{13}{3}\)

c) \(\frac{-4}{13}.\frac{3}{17}+\frac{-12}{13}.\frac{4}{7}+\frac{4}{13}=\frac{4}{13}.\frac{-3}{17}+\frac{4}{13}.\frac{-12}{17}+\frac{4}{13}.1\)

\(=\frac{4}{13}\left(\frac{-3}{17}+\frac{-12}{17}+1\right)=\frac{4}{13}\left(\frac{-15}{17}+\frac{17}{17}\right)=\frac{4}{13}.\frac{2}{17}=\frac{8}{221}\)

d) \(\frac{10^3+2.5+5^3}{55}=\frac{1000+10+125}{55}=\frac{1135}{55}=\frac{227}{11}\)

\(\frac{72}{55}\)

\(A=\frac{\frac{3}{2}+\frac{2}{5}+\frac{1}{10}}{\frac{3}{2}+\frac{2}{3}+\frac{1}{12}}\)

\(\Rightarrow A=\frac{\frac{15}{10}+\frac{4}{10}+\frac{1}{10}}{\frac{18}{12}+\frac{8}{12}+\frac{1}{12}}=\frac{\frac{20}{10}}{\frac{27}{12}}=\frac{2}{\frac{9}{4}}=2:\frac{9}{4}=2.\frac{4}{9}=\frac{8}{9}\)

! Ko bt có đúng ko nx  @@@

~ Học tốt 

# Chiyuki Fujito

a, Ta có : \(\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)

=> \(\frac{4\left(x+1\right)}{12}+\frac{9\left(2x+1\right)}{12}=\frac{2\left(2x+3\left(x+1\right)\right)}{12}+\frac{7+12x}{12}\)

=> \(4\left(x+1\right)+9\left(2x+1\right)=2\left(2x+3\left(x+1\right)\right)+7+12x\)

=> \(4\left(x+1\right)+9\left(2x+1\right)=2\left(2x+3x+3\right)+7+12x\)

=> \(4x+4+18x+9=4x+6x+6+7+12x\)

=> \(4x+18x-12x-6x-4x=6+7-4-9\)

=> \(0x=0\) ( Luôn đúng với mọi x )

Vậy phương trình có vô số nghiệm .

b, Ta có : \(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)

=> \(\frac{2-x}{2001}+1=\frac{1-x}{2002}+1-\frac{x}{2003}+1\)

=> \(\frac{2-x}{2001}+1=\frac{1-x}{2002}+1+\frac{-x}{2003}+1\)

=> \(\frac{2-x}{2001}+\frac{2001}{2001}=\frac{1-x}{2002}+\frac{2002}{2002}+\frac{-x}{2003}+\frac{2003}{2003}\)

=> \(\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)

=> \(\frac{2003-x}{2001}-\frac{2003-x}{2002}-\frac{2003-x}{2003}=0\)

=> \(\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

=> \(2003-x=0\)

=> \(x=2003\)

Vậy phương trình có tập nghiệm là \(S=\left\{2003\right\}\)