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\(\dfrac{6}{21}-\dfrac{-12}{44}+\dfrac{10}{14}-\dfrac{1}{-4}-\dfrac{-18}{33}\)
\(=\dfrac{2}{7}+\dfrac{5}{7}+\dfrac{3}{11}+\dfrac{1}{4}+\dfrac{6}{11}\)
\(=\dfrac{5}{4}+\dfrac{9}{11}=\dfrac{5\cdot11+9\cdot4}{44}=\dfrac{55+36}{44}=\dfrac{91}{44}\)
\(\dfrac{6}{21}\) - \(\dfrac{-12}{44}\) + \(\dfrac{10}{14}\) - \(\dfrac{1}{-4}\) - \(\dfrac{-18}{33}\)
= \(\dfrac{2}{7}\) + \(\dfrac{3}{11}\) + \(\dfrac{5}{7}\) + \(\dfrac{1}{4}\) + \(\dfrac{6}{11}\)
= (\(\dfrac{2}{7}\) + \(\dfrac{5}{7}\)) + (\(\dfrac{3}{11}\) + \(\dfrac{6}{11}\)) + \(\dfrac{1}{4}\)
= 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
\(E=\dfrac{4}{3}+\dfrac{7}{3^2}+\dfrac{10}{3^3}+....+\dfrac{298}{3^{99}}+\dfrac{301}{3^{100}}\)
\(3E=4+\dfrac{7}{3}+\dfrac{10}{3^2}+....+\dfrac{298}{3^{98}}+\dfrac{301}{3^{99}}\)
\(3E-E=4+\dfrac{3}{3}+\dfrac{3}{3^2}+...+\dfrac{3}{3^{98}}+\dfrac{3}{3^{99}}-\dfrac{301}{3^{100}}\)
\(2E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}+\dfrac{1}{3^{98}}+4-\dfrac{301}{3^{100}}\)
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: \(\dfrac{7}{11}+\dfrac{1}{11}\cdot x=0,3-3\dfrac{2}{5}\)
=>\(\dfrac{x}{11}=0,3-3,4-\dfrac{7}{11}=-\dfrac{411}{110}\)
=>\(x=-\dfrac{411}{110}\cdot11=-\dfrac{411}{10}\)
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
\(189:\left[628-\left(2x-1\right)^2\right]=3^2\cdot7\)
=>\(628-\left(2x-1\right)^2=\dfrac{189}{63}=3\)
=>\(\left(2x-1\right)^2=628-3=625\)
=>\(\left[{}\begin{matrix}2x-1=25\\2x-1=-25\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=26\\2x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=-12\end{matrix}\right.\)
\(189:\left[628-\left(2x-1\right)^2\right]=3^2.7\)
\(\Rightarrow628-\left(2x-1\right)^2=\dfrac{189}{63}\)
\(\Rightarrow628-\left(2x-1\right)^2=3\)
\(\Rightarrow\left(2x-1\right)^2=625\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=25\\2x-1=-25\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=13\\x=-12\end{matrix}\right.\)
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