\(A=\frac{1}{3\cdot10}+\frac{1}{10\cdot17}+\frac{1}{17\cdot24}+...+\frac{1}{73\cdot80}-\frac{1}{2\cdot9}-\frac{1}{9\cdot16}-\frac{1}{16\cdot23}-\frac{1}{23\cdot30}\)

\(A=\frac{1}{7}\left(\frac{7}{3\cdot10}+\frac{7}{10\cdot17}+\frac{7}{17\cdot24}+...+\frac{7}{73\cdot80}-\frac{7}{2\cdot9}-\frac{7}{9\cdot16}-\frac{7}{16\cdot23}-\frac{7}{23\cdot30}\right)\)

\(A=\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{73}-\frac{1}{80}-\frac{1}{2}+\frac{1}{9}-\frac{1}{9}+\frac{1}{16}-...-\frac{1}{23}-\frac{1}{30}\right)\)

\(A=\frac{1}{7}\left(\frac{1}{3}-\frac{1}{80}-\frac{1}{2}-\frac{1}{30}\right)\)

a, |x - 3,5| = 7,5

=> x - 3,5 = 7,5 hoặc x - 3,5 = -7,5

=> x = 11 hoặc x = -4 

vậy_

b, |x + 4/5| - 1/2 = 0

=> |x + 4/5| = 1/2

=> x + 4/5 = 1/2  hoặc x + 4/5 = -1/2

=> x = -7/10 hoặc x = -9/10

vậy_

c, 3,6 - |x - 0,4| = 0

=> |x - 0,4| = 3,6

=> x - 0,4 = 3,6 hoặc x - 0,4 = -3,6

=> x = 4 hoặc x = -3,2

vậy_

a)/x-3,5/=7,5

TH1---------x-3,5=7,5       x=7,5+3,5=11

TH2--------x-3,5=-7,5      x=-7,5+3,5=-4

vậy\(x\in\left\{11;-4\right\}\)

#)Giải :

Ta có : \(b2=2c\Rightarrow\frac{b}{c}=\frac{2}{2}=1\Rightarrow\frac{a}{c}=\frac{b}{c}=1\Rightarrow a=b=c\)

\(\Rightarrow\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\left(=1\right)\left(đpcm\right)\)

\(\frac{x}{12}=\frac{3}{x}\Rightarrow x^2=36\)

\(\Rightarrow x=\pm6\)

\(\frac{x}{12}=\frac{3}{x}\)

\(x\)x \(x\)= 12 . 3

\(x^2\)= 36

\(x^2\)= \(6^2\)

=> \(x\)= 6

\(81^7-27^9-9^{13}\)

\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)

\(=3^{28}-3^{27}-3^{26}\)

\(=3^{26}\left(3^2-3-1\right)\)

\(=3^{26}.5=3^{22}.405⋮405\)