a/\(C=\dfrac{2}{1.7}+\dfrac{2}{7.13}+\dfrac{2}{13.19}+...+\dfrac{2}{1013.1019}\)
\(=\dfrac{1}{3}\left(\dfrac{6}{1.7}+\dfrac{6}{7.13}+\dfrac{6}{13.19}+...+\dfrac{6}{1013.1019}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{1013}-\dfrac{1}{1019}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{1019}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{1018}{1019}\)
\(=\dfrac{1018}{3057}\)
b/\(D=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{2011.2019}\)
\(=\dfrac{7}{8}\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{2011.2019}\right)\)
\(=\dfrac{7}{8}\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{2011}-\dfrac{1}{2019}\right)\)
\(=\dfrac{7}{8}\left(1-\dfrac{1}{2019}\right)\)
\(=\dfrac{7}{8}\cdot\dfrac{2018}{2019}\)
\(=\dfrac{7063}{8076}\)

a) Ta có :

\(C=\dfrac{2}{1.7}+\dfrac{2}{7.13}+\dfrac{2}{13.19}+...+\dfrac{2}{1013.1019}\)

\(\Rightarrow C=\dfrac{1}{3}.\left(\dfrac{6}{1.7}+\dfrac{6}{7.13}+\dfrac{6}{13.19}+...+\dfrac{6}{1013.1019}\right)\)

\(\Rightarrow C=\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{1013}-\dfrac{1}{1019}\right)\)

\(\Rightarrow C=\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{1019}\right)\)

\(\Rightarrow C=\dfrac{1}{3}.\dfrac{1018}{1019}\)

\(\Rightarrow C=\dfrac{1018}{3057}\)

b) Ta có:

\(D=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{2011.2019}\)

\(\Rightarrow D=\dfrac{7}{8}.\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{2011.2019}\right)\)

\(\Rightarrow D=\dfrac{7}{8}.\left(\dfrac{1}{1}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{2011}-\dfrac{1}{2019}\right)\)

\(\Rightarrow D=\dfrac{7}{8}.\left(\dfrac{1}{1}-\dfrac{1}{2019}\right)\)

\(\Rightarrow D=\dfrac{7}{8}.\dfrac{2018}{2019}\)

\(\Rightarrow D=\dfrac{7063}{8076}\)

Ta có: \(\left(\frac{8}{1.9}+\frac{8}{9.17}+\frac{8}{17.25}+...+\frac{8}{49.57}\right)+2\left(x-1\right)=\frac{2x+7}{3}+\frac{5x-8}{4}\)

\(\Leftrightarrow1-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+\frac{1}{17}-\frac{1}{25}+....+\frac{1}{49}-\frac{1}{57}+2x-2=\frac{8x+28+15x-24}{12}\)

\(\Leftrightarrow1-\frac{1}{57}+2x-2=\frac{23x+4}{12}\)

\(\Leftrightarrow2x-\frac{58}{57}=\frac{23x+4}{12}\)

\(\Leftrightarrow24x-\frac{232}{19}=23x+4\)

\(\Leftrightarrow x=\frac{308}{19}\)

thấy chưa huj nãy nói sai mà

Ta có: \(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}=1-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+...+\frac{1}{49}-\frac{1}{57}=1-\frac{1}{57}=\frac{56}{57}\)

Vậy: 56/57 + 58/57 + 2x - 2 = 2x + 7/3 + 5x - 8/4

        2 + 2x - 2 = 2x + 7/3 + 5x - 8/4

   2x = 2x + 7/3 + 5x - 8/4

=> 7/3 + 5x - 8/4 = 0

1/3 + 5x = 0

=> 5x = -1/3

=> x = -1/3 : 5=-1/15

a: \(\Leftrightarrow\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+...+\dfrac{1}{49}-\dfrac{1}{57}\right)+2x-2=\dfrac{2}{3}x+\dfrac{7}{3}+\dfrac{5}{4}x-2\)

\(\Leftrightarrow\dfrac{56}{57}+2x-2=\dfrac{23}{12}x+\dfrac{1}{3}\)

=>1/12x=77/57

=>x=308/19

b: =>(x^2-4)(x^2-10)=72

=>x^4-14x^2+40-72=0

=>x^4-14x^2-32=0

=>(x^2-16)(x^2+2)=0

=>x^2-16=0

=>x^2=16

=>x=4 hoặc x=-4

a; (-2,5.0,38.0,4) - {0,125.3,15.(-8)}

= (-2.5.0,4).0,38 - {-8.0,125.3,15}

= - 1.0,38 - {-1.3,15}

= -0,38 + 3,15

= 2,77

b; {(-20,83).0,2 + (-9,17).0,2 : {2,47.0,5 - (-3,53.0,5}

= {0,2.(-20,83 - 9,17)}: {0,5.(2,47 + 3,53)}

= {- 0,2.30}:{0,5.6}

= - 6:6 : 0,5

= -1 : 0,5

= - 2 

\(A=\frac{4^5.13+32^2.65}{4^4.104}=\frac{2^{10}.13+2^{10}.13.5}{2^8.2^3.13}=\frac{2^{10}.13.\left(1+5\right)}{2^{11}.13}=\frac{6}{2}=3\)

Vậy A = 3

\[\frac{{ - {{11}^5}{{.13}^7}}}{{{{11}^5}{{.13}^8}}} = \frac{{ - 1}}{{13}}\]

\(\frac{3}{7}.\frac{9}{26}-\frac{1}{14}.\frac{1}{13}\)

\(=\frac{27}{182}-\frac{1}{182}\)

\(=\frac{1}{7}\)

\(\frac{3}{5}.\frac{13}{46}-\frac{1}{10}.\frac{16}{23}\)

\(=\frac{6}{10}.\frac{13}{46}-\frac{1}{10}.\frac{16}{23}\)

\(=\frac{1}{10}.\frac{78}{46}-\frac{1}{10}.\frac{16}{32}\)

\(=\frac{1}{10}\left(\frac{78}{46}-\frac{32}{46}\right)\)

\(=\frac{1}{10}.1=\frac{1}{10}\)