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\(\left[\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right].\frac{5\left(2x-1\right)}{4x}\)

=\(\left[\frac{4x^2+4x+1-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\right].\frac{5\left(2x-1\right)}{4x}\)

=\(\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5\left(2x-1\right)}{4x}\)

=\(\frac{10}{2x+1}\)

Bạn đăng ít một thôi!

mk lỡ đăng rồi bạn ạ 

nên các bạn giải bài nào cũng được

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

=> \(\frac{6\left(x+1\right)}{12}+\frac{3\left(x-2\right)}{12}=\frac{12}{12}-\frac{8\left(x-1\right)}{12}\)

=> \(6\left(x+1\right)+3\left(x-2\right)=12-8\left(x-1\right)\)

=> \(6x+6+3x-6=12-8x+8\)

=> \(17x=20\)

=> \(x=\frac{20}{17}\)

b, Ta có : \(\frac{5x-1}{6}+x=\frac{6-x}{4}\)

=> \(\frac{5x-1+6x}{6}=\frac{6-x}{4}\)

=> \(4\left(11x-1\right)=6\left(6-x\right)\)

=> \(44x-4-36+6x=0\)

=> \(\)\(50x=40\)

=> \(x=\frac{4}{5}\)

c, Ta có : \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)

=> \(\frac{20\left(1-2x\right)}{12}+\frac{6x}{12}=\frac{9\left(x-5\right)}{12}-\frac{24}{12}\)

=> \(20\left(1-2x\right)+6x=9\left(x-5\right)-24\)

=> \(20-40x+6x-9x+45+24=0\)

=> \(43x=89\)

=> \(x=\frac{89}{43}\)

hỏi chị google ấy

A= \(\frac{1}{31}.\left[\frac{5}{31}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

= \(\frac{1}{31}.\left(\frac{5}{31}.\frac{17}{2}-\frac{17}{2}.\frac{21}{5}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{5}{31}-\frac{21}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{-626}{155}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{1}{31}.\left(\frac{-5321}{155}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{-5321}{4805}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{-5321}{4805}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{30.31}\)

=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{30}-\frac{1}{31}\)

=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{31}\)

=\(\frac{-5321}{4805}+\frac{30}{31}\)

=\(\frac{-671}{4805}\)

chấm ở câu b)  phép nhân à 

b)-47/8 hoặc là bằng -10,23711618

a. \(\frac{45}{2}\)

b.\(\frac{-80}{27}\)

c.2

a, [ 6 + 1/8 - 1/2 ]: 3/12=

            45/8       : 3/12=45/2

b,4/9.(58/3 - 118/3)=

   4/9. (-20)           =-80/9

c,1/4:1/4-2.(-1/2)=

     1      - (-1)    = 2

1. a) \(\frac{-2}{7}+\frac{15}{23}+\frac{\left(-15\right)}{17}+\frac{4}{19}+\frac{8}{23}\)

    \(=\left(\frac{-2}{7}+\frac{-5}{7}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\)

    \(=\left(-1\right)+1+\frac{4}{19}\)

    \(=0+\frac{4}{19}=\frac{4}{19}\)

b) \(\frac{7}{19}\cdot\frac{8}{11}+\frac{7}{19}\cdot\frac{3}{11}+\frac{12}{19}\)

   \(=\frac{7}{19}\cdot\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{12}{19}\)

   \(=\frac{7}{19}\cdot1+\frac{12}{19}\)

   \(=\frac{7}{19}+\frac{12}{19}=\frac{19}{19}=1\)

2. a) \(\frac{1}{3}+\frac{\left(-2\right)}{16}-\frac{7}{14}\)

\(=\frac{5}{24}-\frac{1}{2}\)

\(=-\frac{7}{24}\)

b) \(11\frac{3}{13}-2\frac{4}{7}+5\frac{3}{13}\)

\(=\left(11-2+5\right)+\frac{3}{13}-\frac{4}{7}+\frac{3}{13}\)

\(=14+\left(-\frac{10}{91}\right)\)

\(=-14\frac{10}{91}\)

c) \(0,7\cdot2\frac{2}{3}\cdot20\cdot0,375\cdot\frac{5}{28}\)

\(=\frac{7}{10}\cdot\frac{8}{3}\cdot20\cdot\frac{3}{8}\cdot\frac{5}{28}\)

\(=\left(\frac{7}{10}\cdot\frac{5}{28}\right)\cdot\left(\frac{8}{3}\cdot\frac{3}{8}\right)\cdot20\)

\(=\frac{1}{8}\cdot1\cdot20\)

\(=\frac{20}{8}=\frac{5}{2}\)

d) \(\frac{6}{7}+\frac{5}{7}:5-\frac{8}{9}\)

\(=\frac{6}{7}+\frac{1}{7}-\frac{8}{9}\)

\(=1-\frac{8}{9}\)

\(=\frac{1}{9}\)

~Học tốt~