\(=\dfrac{7-2}{2.7}+\dfrac{12-7}{7.12}+...+\dfrac{102-97}{97.102}\)

\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{97}-\dfrac{1}{102}\)

\(=\dfrac{1}{2}-\dfrac{1}{102}=\dfrac{25}{51}\)

\(=\dfrac{7-2}{2.7}+\dfrac{12-7}{7.12}+...+\dfrac{97-92}{92.97}\)

\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{92}-\dfrac{1}{97}\)

\(=\dfrac{1}{2}-\dfrac{1}{97}=\dfrac{95}{194}\)

Mình xin được sửa lại vì câu đầu mình nhầm đề bài ạ

a) \(215+x=400\)

\(\Rightarrow x=400-215\)

\(\Rightarrow x=185\)

b) \(12,5-2x=\dfrac{1}{5}\)

\(\Rightarrow2x=12,5-\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{123}{10}\)

\(\Rightarrow x=\dfrac{123}{10}:2\)

\(\Rightarrow x=\dfrac{123}{20}\)

còn câu c d nhanh cậu ơi

Bạn xem lại đề phần e nha .

\(c)x.0,25-\dfrac{3}{25}:x=2\dfrac{1}{4}\)

\(x.0,25-\dfrac{25}{3}.x=\dfrac{9}{4}\)

\(x.\left(0,25-\dfrac{25}{3}\right)=\dfrac{9}{4}\)

\(x.\dfrac{-97}{12}=\dfrac{9}{4}\)

\(x=\dfrac{-27}{97}\)

\(d)\left(x-\dfrac{1}{3}\right).\left(x-\dfrac{1}{4}\right)=0\)

Do : 2 số đó nhân với nhau bằng 0 nên một trong 2 số là số 0.

Do đó :

\(Th1:x-\dfrac{1}{3}=0\)

\(x=\dfrac{1}{3}\)

\(Th2:x-\dfrac{1}{4}=0\)

\(x=\dfrac{1}{4}\)

`2x-2/3=1/2`

`2x=1/2+2/3`

`2x=7/6`

`x=7/6:2=7/12`

\(2x-\dfrac{2}{3}=\dfrac{1}{2}\Leftrightarrow2x=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\Leftrightarrow x=\dfrac{7}{6}:2=\dfrac{7}{12}\)

2x - 2/3 = 1/2

2x = 1/2 + 2/3 = 7/6

x = 7/6 : 2 = 7/12

vậy x = 7/12

Em cần phần nào nhỉ .

A = \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+\(\dfrac{5}{16.21}\)+...+\(\dfrac{5}{101.106}\)

A = \(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{106}\)

A = \(\dfrac{105}{106}\)

B = \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)

B = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)

B = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)

B = \(\dfrac{99}{100}\)

C = \(\dfrac{1}{2.7}+\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{97.102}\)

C= \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{97.102}\))

C = \(\dfrac{1}{5}\)\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\)+...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{102}\))

C = \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{102}\))

C = \(\dfrac{1}{5}\) \(\times\) \(\dfrac{25}{51}\)

C = \(\dfrac{5}{51}\) 

D = \(\dfrac{1}{2}\) +   \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)

D = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+ \(\dfrac{1}{8.9}\)

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{9}\)

D = \(\dfrac{8}{9}\)

E = \(\dfrac{3}{2.4}\)+\(\dfrac{3}{4.6}\)+\(\dfrac{3}{6.8}\)+...+\(\dfrac{3}{98.100}\)

E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))

E = \(\dfrac{3}{2}\)\(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\))

E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))

E = \(\dfrac{3}{2}\) \(\times\) \(\dfrac{49}{100}\)

E = \(\dfrac{147}{200}\)

\(4\times x^3+15=47\)

\(4\times x^3=32\)

\(x^3=8\)

\(x=2\)

toán lp 5 ak bn ? dễ quá ha --_--''

a) 4.x³+15=47

=> 4.x³=47-15

=> x³=32:4

=> x³=8

=> x³=2³

=> x=2

b) 4.2^x-3=125

=> 4.2^x=125+3

=> 2^x=128:4

=> 2^x=32

=> 2^x=2⁵

=> x=5

Đọc đề hơi băn khoăn. $2 x 5$ là $\overline{2x5}$ hả bạn?

X= 37 + 2 x 5

x=37+10

x=47

\(=5+\dfrac{4}{5}\left(\dfrac{5}{2\cdot7}+\dfrac{5}{7\cdot12}+...+\dfrac{5}{52\cdot57}\right)\)

\(=5+\dfrac{4}{5}\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{52}-\dfrac{1}{57}\right)\)

\(=5+\dfrac{4}{5}\cdot\dfrac{55}{114}=\dfrac{307}{57}\)

\(a,\dfrac{7}{12}-\left(x+\dfrac{7}{10}\right):\dfrac{6}{5}=\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{7}{12}-x-\dfrac{7}{10}:\dfrac{6}{5}=\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{7}{12}-x-\dfrac{7}{12}=\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{7}{12}-x=\dfrac{5}{4}+\dfrac{7}{12}\)

\(\Leftrightarrow\dfrac{7}{12}-x=\dfrac{11}{6}\)

\(\Leftrightarrow x=\dfrac{7}{12}-\dfrac{11}{6}\)

\(\Leftrightarrow\dfrac{-5}{4}\)

tham khảo !