\(\frac{-8}{10}.\frac{74}{7}+\frac{17}{7}.\frac{-8}{13}+\frac{3}{5}=\frac{-8}{7}.\frac{37}{5}+\frac{-8}{7}.\frac{17}{13}+\frac{3}{5}\)

\(=\frac{-8}{7}.\left(\frac{37}{5}+\frac{17}{13}\right)+\frac{3}{5}=\frac{-8}{7}.\frac{566}{65}+\frac{3}{5}\)

\(=\frac{-4528}{455}+\frac{273}{455}=\frac{-4255}{455}=\frac{851}{91}\)

851/91

\(2C=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{38.39}\)
\(C=\frac{617}{1482}\)

\(3D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3D-D=1-\frac{1}{3^8}\)
\(D=\frac{1}{2}-\frac{1}{2.3^8}\)

Ta có:\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{38.39}\right)\)

b,\(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)

\(\Rightarrow3D=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^7}\)

\(\Rightarrow2D=1-\frac{1}{3^8}\)

\(\Rightarrow D=\frac{3^8-1}{3^8}:2\)

a)\(=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)

\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\left(\frac{15}{26}+\frac{2}{13}\right)\)

\(=0-\frac{19}{26}\)

\(=-\frac{19}{26}\)

c)\(=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)

\(=\frac{-11}{23}.2-\frac{1}{23}\)

\(=\frac{-22}{23}-\frac{1}{23}\)

\(=-1\)

Lời giải:

a)

\(-3\frac{5}{8}+\left(-\frac{3}{8}+\frac{9}{4}\right)\)

\(=-\frac{29}{8}+\left(-\frac{3}{8}+\frac{18}{8}\right)\)

\(=-\frac{29}{8}+\frac{15}{8}=-\frac{14}{8}=-\frac{7}{4}\)

b) \(\frac{\left(-9\right)\cdot11+32\cdot\left(-9\right)}{\left(-43\right)\cdot15+12\cdot\left(-43\right)}=\frac{\left(-9\right)\left(11+32\right)}{\left(-43\right)\left(15+12\right)}=\frac{\left(-9\right)\cdot43}{\left(-43\right)\cdot27}=\frac{\left(-1\right)\cdot1}{\left(-1\right)\cdot3}=\frac{1}{3}\)

c) Thay \(x=\frac{2011}{2012}\)vào biểu thức \(x\cdot\frac{1}{3}+2x\cdot\frac{3}{6}-3x\cdot\frac{4}{9}\)ta có :

\(\frac{2011}{2012}\cdot\frac{1}{3}+2\cdot\frac{2011}{2012}\cdot\frac{3}{6}-3\cdot\frac{2011}{2012}\cdot\frac{4}{9}\)

\(=\frac{2011}{2012}\cdot\frac{1}{3}+2\cdot\frac{2011}{2012}\cdot\frac{1}{2}-3\cdot\frac{2011}{2012}\cdot\frac{4}{9}\)

\(=\frac{2011}{6036}+\frac{2011}{2012}-\frac{2011}{1509}\)

\(=\frac{2011}{6036}+\frac{6033}{6036}-\frac{8044}{6036}=\frac{2011+6033-8044}{6036}=0\)

kết quả bằng 9/5 đó bạn

giải ra các cách được hông bạn

a) \(\frac{27}{8}\times\frac{5}{7}\times\frac{8}{27}+\frac{2}{7}\)

\(=\frac{27\times5\times8}{8\times7\times27}+\frac{2}{7}\)

\(=\frac{5}{7}+\frac{2}{7}\)

\(=1\)

b) \(1,6\times\frac{25}{32}-\left(\frac{2}{3}+\frac{4}{5}\right):\frac{11}{5}\)

\(=\frac{8}{5}\times\frac{25}{32}-\left(\frac{2}{3}+\frac{4}{5}\right).\frac{5}{11}\)

\(=\frac{5}{4}-(\frac{10}{33}+\frac{4}{11})\)

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

\(=\frac{7}{12}\)

chịu luôn rùi! !   phần b, mk ko tính nhanh đc! cho mk xin lỗi bn nhiều nha!

bài 1 a)

\(\frac{15}{7}\times\frac43+\frac63\)

=\(\frac{60}{21}+\frac63\)

=\(\frac{60}{21}+\frac{42}{21}\)

=\(\frac{102}{21}\)

=\(\frac{34}{7}\)

b)\(\frac{-13}{41}\times\frac{17}{11}\times\frac{11}{17}\)

=\(\frac{-13}{41}\times\left(\frac{17}{11}\times\frac{11}{17}\right)\)

=\(\frac{-13}{41}\times1\)

=\(\frac{-13}{41}\)

a ) Co :

 1/1.2 - 1/2.3 = 2/1.2.3 

 1/2.3 - 1/3.4 = 2/2.3.4

 ...

 1/37.38 - 1/38.39 = 2/37.38.39

=> 2M = 2/1.2.3 + 2/2.3.4 + ... + 2/37.38.39

=> 2M = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ... + 1/37.38 - 1/38.39

=> 2M = 1/2 - 1/1482

=> 2M = 370/741

=> M = 185/741

B ) A = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^8

3A = 1 + 1/3 + 1/3^2 + ... + 1/3^7

3A - A = ( 1 + 1/3 + 1/3^2 + ... + 1/3^7 ) - ( 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^8 )

2A = 1 - 1/3^8

A = ( 1 - 1/3^8 ) / 2