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

\(189:\left[628-\left(2x-1\right)^2\right]=3^2\cdot7\)

=>\(628-\left(2x-1\right)^2=\dfrac{189}{63}=3\)

=>\(\left(2x-1\right)^2=628-3=625\)

=>\(\left[{}\begin{matrix}2x-1=25\\2x-1=-25\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=26\\2x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=-12\end{matrix}\right.\)

\(189:\left[628-\left(2x-1\right)^2\right]=3^2.7\)
\(\Rightarrow628-\left(2x-1\right)^2=\dfrac{189}{63}\)
\(\Rightarrow628-\left(2x-1\right)^2=3\)
\(\Rightarrow\left(2x-1\right)^2=625\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=25\\2x-1=-25\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=13\\x=-12\end{matrix}\right.\)

\(C=\dfrac{4}{3.5}+\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{97.99}\)
\(=2\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(=2\cdot\dfrac{32}{99}=\dfrac{64}{99}\)
\(D=\dfrac{18}{2.5}+\dfrac{18}{5.8}+...+\dfrac{18}{203.206}\)
\(=6\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{203.206}\right)\)
\(=6\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{203}-\dfrac{1}{206}\right)\)
\(=6\left(\dfrac{1}{2}-\dfrac{1}{206}\right)\)
\(=6\cdot\dfrac{51}{103}=\dfrac{306}{103}\)
\(\Rightarrow\dfrac{C}{D}=\dfrac{\dfrac{64}{99}}{\dfrac{306}{103}}=\dfrac{3296}{15147}\)

ai trả lời giúp mik với

a: Hai tia trùng nhau là OA,OB

Hai tia đối nhau là Bx,BO; Ax;AO

b: Trên tia Ox, ta có: OA<OB

nên A nằm giữa O và B

c: Ta có: A nằm giữa O và B

=>OA+AB=OB

=>AB+2=4

=>AB=2(cm)

ta có: A nằm giữa O và B

mà AO=AB(=2cm)

nên A là trung điểm của OB

a: Hai tia trùng nhau là OA,OB

Hai tia đối nhau là Bx,BO; Ax;AO

b: Trên tia Ox, ta có: OA<OB

nên A nằm giữa O và B

c: Ta có: A nằm giữa O và B

=>OA+AB=OB

=>AB+2=4

=>AB=2(cm)

ta có: A nằm giữa O và B

mà AO=AB(=2cm)

nên A là trung điểm của OB