,làm ơn giúp mik với ah

\(\left(1+\dfrac{2}{3}\right).\left(1+\dfrac{2}{4}\right).\left(1+\dfrac{2}{5}\right)....\left(1+\dfrac{2}{2020}\right).\left(1+\dfrac{2}{2021}\right)\)

= \(\dfrac{5}{3}.\dfrac{6}{4}.\dfrac{7}{5}.\dfrac{8}{6}.\dfrac{9}{7}....\dfrac{2022}{2020}.\dfrac{2023}{2021}\)

= \(\dfrac{1}{3}.\dfrac{1}{4}.2022.2023\)

= \(\dfrac{337.2023}{2}\)

= \(\dfrac{\text{681751}}{2}\)

chịu

ăn ba tô cơm

\(4^{2020}+4^{2020}+4^{2020}+4^{2020}+4^{2020}\\=5\cdot4^{2020}\)

Để yêu cầu gì thế bạn?

\((\frac{4}{3}-\frac{1}{4}-\frac{5}{12})\)+2x=\(\frac{8}{5}:\frac{3}{5}\)

=\(\frac{2}{3}\)+2x=\(\frac{8}{3}\)

2x=\(\frac{8}{3}-\frac{2}{3}\)

2x=2

x=2:2

x=1

Vậy x=1

\(\left(\frac{4}{3}-\frac{1}{4}-\frac{5}{12}\right)+2x=\frac{8}{5}:\frac{3}{5}\)

 \(\left(\frac{16}{12}-\frac{3}{12}-\frac{5}{12}\right)+2x=\frac{8}{5}.\frac{5}{3}\)

          \(\frac{2}{3}+2x=\frac{8}{3}\)

                     \(2x=\frac{8}{3}-\frac{2}{3}\) 

                      \(2x=2\)

                           \(x=2:2\)

                              \(x=1\)

                       Vậy \(x=1\)

Chúc bạn học thật tốt !!!

Ta có

        4/3 - 1/4 - 5/12 + 2x = 8/5 : 3/5

        => 2/3 + 2x = 8/3

         =>2x=2=>x=1

a)=<

b)=-1/6<1/12

  1-2+3-4+5-6+...+99-100+101 
= (1+3+5+...+101) - (2+4+6+...+100) 
tu 1 den 101 co : (101-1):2+1=51 
1+..+101 = (1+101)x 51:2= 2601 
tu 2 den 100 co : (100-2);2+1=50 
2+...+100 = (100 +2) x 50:2=2550 
=> A= 2601-2550=51

học tốt

tích nha,mơn nhìu

thanks nhìu

=9 mũ 2021 

\(F=9^2+9^3+9^4+...+9^{2020}\)

\(\Rightarrow9F=9^3+9^4+9^5+...+9^{2021}\)

\(\Rightarrow9F-F=\left(9^3+9^4+9^5+...+9^{2021}\right)-\left(9^2+9^3+9^4+...+9^{2020}\right)\)

\(\Rightarrow8F=9^{2021}-9^2\)

\(\Rightarrow F=\frac{9^{2021}-9^2}{8}\)