mk ko viết lại đề

\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}+\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)

\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}+\frac{2^{12}.3^{10}\left(1+5\right)}{2.\left(2^{12}.3^{12}\right)}\)

\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)

Vậy A= \(\frac{1}{2}\)

=> 1+2+3+4+5+....+x = 190

x(x+1) = 190.2 = 380

x(x+1) = 19.(19 + 1)

VẬy x = 19 

\(2+2^2+...+2^{100}\\ =\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\\ =2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\\ =\left(1+2\right)\left(2+2^3+...+2^{99}\right)\\ =3\left(2+2^3+...+2^{99}\right)⋮3\)

19 nha bn

3³ đâu mất r????

Đề hơi khó hiểu bn nhé

\(\left(\dfrac{1}{4}x-\dfrac{1}{8}\right).\dfrac{3}{4}=\dfrac{1}{4}\)

\(\dfrac{1}{4}x-\dfrac{1}{8}=\dfrac{1}{4}:\dfrac{3}{4}=\dfrac{1}{3}\)

\(\dfrac{1}{4}x=\dfrac{1}{3}+\dfrac{1}{8}=\dfrac{11}{24}\)

\(x=\dfrac{11}{24}:\dfrac{1}{4}=\dfrac{11}{6}\)

\(a,x+\dfrac{3}{7}=\dfrac{2}{5}+\dfrac{3}{10}\)

\(x+\dfrac{3}{7}=\dfrac{4}{10}+\dfrac{3}{10}\)

\(x+\dfrac{3}{7}=\dfrac{7}{10}\)

\(x=\dfrac{7}{10}-\dfrac{3}{7}\)

\(x=\dfrac{49}{70}-\dfrac{30}{70}\)

\(x=\dfrac{19}{70}\)

\(b,\dfrac{19}{20}-x=\dfrac{8}{5}-\dfrac{3}{4}\)

\(\dfrac{19}{20}-x=\dfrac{32}{20}-\dfrac{15}{20}\)

\(\dfrac{19}{20}-x=\dfrac{17}{20}\)

\(x=\dfrac{19}{20}-\dfrac{17}{20}\)

\(x=\dfrac{2}{20}\)

\(x=\dfrac{1}{10}\)

#Urushi☕

=>3^{1+2+3+4+...+x}=3¹⁹⁰

=>1+2+3+4+...+x=190

=>(x+1).x:2=190

=>(x+1).x=380

mà 380 chỉ có thể ptích thành: 380=20.19

=>x+1=20 và x=19

vậy x=19

S= 3.(1+2+3+...+50)

S= 3.(1+2+3+...+50)

S= 3.1275

S= 3825