\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)

\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)

\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3x+2}-\frac{1}{3x+5}\right)=\frac{4}{25}\)

\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3x+5}\right)=\frac{4}{25}\)

\(\frac{1}{2}-\frac{1}{3x+5}=\frac{12}{25}\)

\(\frac{1}{3x+5}=\frac{1}{50}\)

=> 3x+5 = 50

3x = 45

x = 15

\(x\) \(\times\) \(\dfrac{4}{5}\) = 15

\(x\)         = 15 : \(\dfrac{4}{5}\)

\(x\)         = \(\dfrac{75}{4}\) 

\(\dfrac{12}{25}\) \(\times\) \(x\) = \(\dfrac{4}{10}\)

          \(x\) = \(\dfrac{4}{10}\) : \(\dfrac{12}{25}\)

          \(x\) = \(\dfrac{5}{6}\) 

\(\dfrac{6}{5}\): \(x\) = \(\dfrac{1}{3}\)\(x\) : \(\dfrac{2}{7}\) = \(\dfrac{3}{5}\) (xem lại đề bài)

1) \(\left(5x-1\right)\left(5x+1\right)=25x^2-7x+15\)

\(\Leftrightarrow25x^2-1=25x^2-7x+15\)

\(\Leftrightarrow7x=16\Leftrightarrow x=\dfrac{16}{7}\)

2) \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)

\(\Leftrightarrow3x^2-2x-5-3x^2-2x+1=x-4\)

\(\Leftrightarrow5x=0\Leftrightarrow x=0\)

a/ (3x - 1).(1/2.5) = 0 => 3x - 1 = 0 => 3x = 1 => x = 1/3

b/ 1/4 + 1/3 : (2x - 1) = 5 => 1/3 : (2x - 1) = 19/4 => 2x - 1 = 4/57 => 2x = 61/57 => x = 61/114

c/ (2x + 2/5)² - 9/25 = 0 => (2x + 2/5)² = 9/25 => 2x + 2/5 = 3/5 => 2x = 1/5 => x = 1/10 

                                                         hoặc             2x + 2/5 = -3/5 => 2x = -1 => x = -1/2

     Vậy x = {1/10 ; -1/2}

d/ (3x - 1/2)³ + 1/9 = 0 => (3x - 1/2)³ = -1/9 => 3x - 1/2 = -1/3 => 3x = 1/6 => x = 1/18

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