\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)

=>\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{23}{27}-\dfrac{5}{9}=\dfrac{8}{27}\)

=>\(2+\dfrac{3}{4}x=\dfrac{7}{9}:\dfrac{8}{27}=\dfrac{7}{9}\cdot\dfrac{27}{8}=\dfrac{21}{8}\)

=>\(\dfrac{3}{4}x=\dfrac{21}{8}-2=\dfrac{5}{8}\)

=>\(x=\dfrac{5}{8}:\dfrac{3}{4}=\dfrac{5}{8}\cdot\dfrac{4}{3}=\dfrac{5}{6}\)

\(\left(x-\dfrac{1}{3}\right):\dfrac{1}{2}+\dfrac{3}{4}=\dfrac{38}{7}\)

=>\(\left(x-\dfrac{1}{3}\right):\dfrac{1}{2}=\dfrac{38}{7}-\dfrac{3}{4}=\dfrac{152-21}{28}=\dfrac{131}{28}\)

=>\(x-\dfrac{1}{3}=\dfrac{1}{2}\cdot\dfrac{131}{28}=\dfrac{131}{56}\)

=>\(x=\dfrac{131}{56}+\dfrac{1}{3}=\dfrac{449}{168}\)

Câu hỏi đâu bạn nhỉ?

Bạn cần giúp gì nhỉ?

\(\dfrac{6}{21}-\dfrac{-12}{44}+\dfrac{10}{14}-\dfrac{1}{-4}-\dfrac{-18}{33}\)

\(=\dfrac{2}{7}+\dfrac{5}{7}+\dfrac{3}{11}+\dfrac{1}{4}+\dfrac{6}{11}\)

\(=\dfrac{5}{4}+\dfrac{9}{11}=\dfrac{5\cdot11+9\cdot4}{44}=\dfrac{55+36}{44}=\dfrac{91}{44}\)

   \(\dfrac{6}{21}\) - \(\dfrac{-12}{44}\) + \(\dfrac{10}{14}\) - \(\dfrac{1}{-4}\) - \(\dfrac{-18}{33}\)

= \(\dfrac{2}{7}\) + \(\dfrac{3}{11}\) + \(\dfrac{5}{7}\) + \(\dfrac{1}{4}\) + \(\dfrac{6}{11}\)

= (\(\dfrac{2}{7}\) + \(\dfrac{5}{7}\)) + (\(\dfrac{3}{11}\) + \(\dfrac{6}{11}\))  + \(\dfrac{1}{4}\)

= 1 + \(\dfrac{9}{11}\) + \(\dfrac{1}{4}\)

 = \(\dfrac{20}{11}\) + \(\dfrac{1}{4}\)

= \(\dfrac{91}{44}\)

a: Số lần số chấm xuất hiện là số chẵn là:

21+29+25=75(lần)

=>Xác suất là \(\dfrac{75}{120}=\dfrac{5}{8}\)

b: Số lần xuất hiện số chấm là số chia hết cho 2 là:

21+29+25=75(lần)

=>Xác suất là \(\dfrac{75}{120}=\dfrac{5}{8}\)

c: Số lần xuất hiện số chấm là số nguyên tố là:

21+12+16=49(lần)

=>Xác suất là \(\dfrac{49}{120}\)

d: Số lần xuất hiện số chấm là số lẻ:

17+12+16=45(lần)

=>Xác suất là \(\dfrac{45}{120}=\dfrac{3}{8}\)

(\(x-\dfrac{1}{3}\)) : \(\dfrac{1}{2}\) + \(\dfrac{3}{7}\) = 5\(\dfrac{3}{7}\)

(\(x-\dfrac{1}{3}\)) : \(\dfrac{1}{2}\)  + \(\dfrac{3}{7}\) = \(\dfrac{38}{7}\)

(\(x\) - \(\dfrac{1}{3}\)) : \(\dfrac{1}{2}\)           = \(\dfrac{38}{7}\) - \(\dfrac{3}{7}\)

(\(x-\dfrac{1}{3}\)) : \(\dfrac{1}{2}\)         = 5

\(x\) - \(\dfrac{1}{3}\)                   = 5 x \(\dfrac{1}{2}\)

\(x\) - \(\dfrac{1}{3}\)                  = \(\dfrac{5}{2}\)

\(x\)                         = \(\dfrac{5}{2}\) + \(\dfrac{1}{3}\)

\(x\)                        =  \(\dfrac{17}{6}\)

Vậy \(x=\dfrac{17}{6}\)

-2/3x +1/5x = -14/15

-7/15x -14/15

x = (-14/15) : ( -7/15)

x = 2

-7/15x = -14/15 nha, mình ghi sai 

   \(E=\dfrac{4}{3}+\dfrac{7}{3^2}+\dfrac{10}{3^3}+....+\dfrac{298}{3^{99}}+\dfrac{301}{3^{100}}\)

\(3E=4+\dfrac{7}{3}+\dfrac{10}{3^2}+....+\dfrac{298}{3^{98}}+\dfrac{301}{3^{99}}\)

\(3E-E=4+\dfrac{3}{3}+\dfrac{3}{3^2}+...+\dfrac{3}{3^{98}}+\dfrac{3}{3^{99}}-\dfrac{301}{3^{100}}\)

\(2E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}+\dfrac{1}{3^{98}}+4-\dfrac{301}{3^{100}}\)

Đặt

\(A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{97}}+\dfrac{1}{3^{98}}\)

\(3A=3+1+\dfrac{1}{3}+....+\dfrac{1}{3^{96}}+\dfrac{1}{3^{97}}\)

\(3A-A=3-\dfrac{1}{3^{98}}\)

\(2A=3-\dfrac{1}{3^{98}}\)

\(A=\dfrac{3}{2}-\dfrac{1}{3^{98}\times2}\)

\(\Rightarrow2E=\dfrac{3}{2}-\dfrac{1}{3^{98}\times2}+4-\dfrac{301}{3^{100}}\)

      \(2E=\dfrac{11}{2}-\dfrac{1}{3^{98}\times2}-\dfrac{301}{3^{100}}\) 

\(\Rightarrow2E< \dfrac{11}{2}\Rightarrow E< \dfrac{11}{4}=2,75\)

a: \(\dfrac{7}{2}-1,5+\dfrac{-8}{11}=3,5-1,5-\dfrac{8}{11}=2-\dfrac{8}{11}=\dfrac{22-8}{11}=\dfrac{14}{11}\)

b: \(\left(1\dfrac{3}{7}+3\dfrac{4}{7}\right):\dfrac{6}{5}-2,5\cdot\dfrac{8}{15}\)

\(=\left(1+\dfrac{3}{7}+3+\dfrac{4}{7}\right)\cdot\dfrac{5}{6}-\dfrac{5}{2}\cdot\dfrac{8}{15}\)

\(=5\cdot\dfrac{5}{6}-\dfrac{40}{30}=\dfrac{25}{6}-\dfrac{4}{3}=\dfrac{17}{6}\)

c: \(\dfrac{7}{9}\cdot\dfrac{35}{17}-\dfrac{7}{9}\cdot\dfrac{25}{17}-\dfrac{7}{9}\)

\(=\dfrac{7}{9}\left(\dfrac{35}{17}-\dfrac{25}{17}-1\right)\)

\(=\dfrac{7}{9}\cdot\dfrac{-7}{17}=\dfrac{-49}{117}\)

d: \(\left(120\%+1\dfrac{3}{5}\right):\dfrac{27}{5}-4,5\cdot\dfrac{4}{9}\)

\(=\left(1,2+1,6\right)\cdot\dfrac{5}{27}-2\)

\(=\dfrac{14}{27}-2=\dfrac{14}{27}-\dfrac{54}{27}=-\dfrac{40}{27}\)

e: \(\dfrac{-5}{7}\cdot x=\dfrac{25}{12}\)

=>\(x=-\dfrac{25}{12}:\dfrac{5}{7}=-\dfrac{25}{12}\cdot\dfrac{7}{5}=\dfrac{-5\cdot7}{12}=-\dfrac{35}{12}\)

g: \(\dfrac{7}{11}+\dfrac{1}{11}\cdot x=0,3-3\dfrac{2}{5}\)

=>\(\dfrac{x}{11}=0,3-3,4-\dfrac{7}{11}=-\dfrac{411}{110}\)

=>\(x=-\dfrac{411}{110}\cdot11=-\dfrac{411}{10}\)

hơi dài đó.

a: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2021}{2022}\)

=>\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)

=>\(1-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)

=>\(\dfrac{1}{x+1}=1-\dfrac{2021}{2022}=\dfrac{1}{2022}\)

=>x+1=2022

=>x=2021

b: Sửa đề: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

=>\(\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)=0\)

=>\(\dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\)

=>\(\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)

=>x+100=0

=>x=-100

ai trả lời giúp mik đi mà