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a) Ta có B = \(\left(\frac{2}{15}+\frac{3}{40}+\frac{4}{96}+\frac{5}{204}+\frac{6}{391}\right).x.\left(x-1\right)=\frac{20}{69}\)

         => \(\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}+\frac{6}{17.23}\right).x.\left(x-1\right)=\frac{20}{69}\)

         => \(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}\right).x.\left(x-1\right)=\frac{20}{69}\)

        => \(\left(\frac{1}{3}-\frac{1}{23}\right).x.\left(x-1\right)=\frac{20}{69}\)

        => \(\frac{20}{69}.x.\left(x-1\right)=\frac{20}{69}\)

       => \(x.\left(x-1\right)=\frac{20}{69}:\frac{20}{69}\)

      => \(x.\left(x-1\right)=1\)

      => \(x\in\varnothing\) 

a)  \(\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+....+\frac{1}{8554}\right).x=\frac{31}{94}\)

=> \(\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{91.94}\right).x=\frac{31}{94}\) 

=> \(\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}\right)=\frac{31}{94}\)

=> \(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{91}-\frac{1}{94}\right).x=\frac{31}{94}\)

=> \(\frac{1}{3}.\left(1-\frac{1}{94}\right).x=\frac{31}{94}\)

=> \(\frac{1}{3}.\frac{93}{94}.x=\frac{31}{94}\)

=> \(\frac{31}{94}.x=\frac{31}{94}\)

=> \(x=\frac{31}{94}:\frac{31}{94}\)

=> \(x=1\)

pt đã cho có dạng \(\frac{1}{\left(x+1\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+10\right)}=\frac{4}{13}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+10}=\frac{4}{13}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{4}{13}\Leftrightarrow....\)

bạn tuấn mình thấy vậy nè

Gỉa sử cho x=1 ta thấy \(\frac{1}{1\times4}\ne\frac{1}{1}-\frac{1}{4}\)

Bạn bấm máy tính thử xem dấu bằng chỉ áp dụng với 2 số tự nhiên liên tiếp thôi còn cái này cách 3 lận

giải thích giúp mình với

à 

ý là nhân hai về pt cho 3 đúg kk 

mình hiểu rồi nha

cảm ơn nha

\(\frac{3}{1.4}+\frac{3}{4.7}+..+\frac{3}{97.100}=\frac{0,33x}{2009}\)

\(1-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{100}=\frac{0,33x}{2009}\)

\(1-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{99}{100}=\frac{0,33x}{20009}\Rightarrow2009.99=100.0,33x\)

x=6027

\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{9700}=\frac{0,33x}{2009}\)

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=\frac{0,33x}{2009}\)

\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{1}{1}-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{100}{100}-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{99}{100}=\frac{0,33x}{2009}\)

\(\Rightarrow2009.99=100.0,33x\)

\(\Rightarrow2009.99=33x\)

\(\Rightarrow2009.99:33=x\)

\(\Rightarrow2009.3=x\)

\(\Rightarrow6027=x\)

Vậy \(x=6027\)(MK KO CHẮC NÓ ĐÚNG NHÉ )

a)    (2x + 1)(3x - 2) = (5x - 8)(2x + 1)

 <=> 6x² - x - 2 = 10x² - 11x - 8

<=>  6x² - 10x² - x + 11x -2 + 8 = 0

<=>  -4x² + 10x + 6  = 0

<=> -2 (2x² - 5x - 3) = 0

<=> 2x² - 5x - 3 = 0 

<=> 2x² - 6x + x - 3 = 0

<=> x (2x + 1) - 3 (2x + 1) = 0

<=> (x - 3) (2x + 1) = 0

* x - 3 = 0  => x = 3

* 2x + 1 = 0 => x = -1/2 

S = {-1/2; 3}

b) 4x² – 1 = (2x +1)(3x -5)

<=> 4x² – 1 - (2x +1)(3x -5) = 0

<=> (2x - 1) (2x + 1) - (2x + 1)(3x - 5) = 0

<=>  (2x + 1) (2x - 1 - 3x + 5) = 0

<=>  (2x + 1) (-x + 4) = 0

* 2x + 1 = 0  <=> x = -1/2

* -x + 4 = 0 <=> x = 4

S = {-1/2; 4}

c) (x + 1)² = 4(x² – 2x + 1)

<=> (x + 1)² - 4(x² – 2x + 1) = 0

<=> (x + 1)² - 4(x² – 1)² = 0

* (x + 1)² = 0   <=> x = -1

* 4(x² - 1)² = 0  <=> x = 1 và x = -1

S = {-1;  1}

d) 2x³ + 5x² – 3x = 0

<=> x (2x² + 5x - 3) = 0

<=> x (2x² + 6x - x - 3) = 0

<=> x [x(2x - 1) + 3 (2x - 1)] = 0

<=> x (2x - 1) (x + 3) = 0

* x = 0

* 2x - 1 = 0  <=> x = 1/2

* x + 3 = 0  <=> x = -3

S = { -3; 0; 1/2}

\(\frac{1}{x^2+5x+4}+\frac{1}{x^2+11x+28}+\frac{1}{x^2+17x+70}=\frac{3}{4x-2}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+10\right)}=\frac{3}{4x-2}\)

\(\Leftrightarrow3x^2+21x+36=0\)

\(\Leftrightarrow x=-3\)

A = 1/4 + 1/28 + 1/70 +...+ 1/9700

A = 1/1.4 + 1/4.7 + 1/7.10 +...+ 1/97.100

3A = 3/1.4 + 3/4.7 + 3/7.10 +...+ 3/97.100

3A = 1 - 1/100

3A = 99/100

A=99/100:3=33/100

\(=\frac{1}{1.4}+\frac{1}{4.7}+..+\frac{1}{97.100}\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)

= \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}\)

= \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)

= \(\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

= \(\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{100}\right)\)

= \(\frac{1}{3}.\frac{99}{100}\)

=\(\frac{33}{100}\)

bằng \(\frac{539}{1080}\)

= 539/1080

câu trả lời là \(\frac{539}{1080}\)