ta có :

= ( 1 + 59049 ) + ( 3 + 2187 ) + ( 9 + 6561 ) + ( 27 + 243 ) + ( 81 + 729 )

= 59050 + 2190 + 6570 + 270 + 810

= 59050 + ( 2190 + 810 ) + 6570 + 270

= 59050 + 3000 + 6570 + 270

= 59050 + ( 3000 + 6570 ) + 270

= 59050 + 9570 + 270

= 68620 + 270

= 68890

68890

Kết quả là 68890

Nhớ trả lời cho mình

🐰

Bài làm

    1 + 3 + 9 + 27 + 6561 + 19683

= ( 1 + 9 ) + ( 3 + 27 ) + ( 6561 + 19683 )

= 10 + 30 +  26244

= 40 + 26244

= 26284

Thanks

                  Bài giải

\(=1+3+\left(9+6561\right)+\left(27+19683\right)\)

\(=1+3+6570+19710\)

\(=26284.\)

cái này là toán lớp 6 mak!câu c có số nguyên âm nè!!!!!!!!!!

Chịu nha!!!!Mình mới lớp 5 thôi.Sorry nha

b,(1+2+3+.....) *(25*3-75) +75=(1+2+3+...)*(75-75)+75=(1+2+3+....)*0+75=0+75=75

1+3+9+27+....+2187+6561

đặt A = 1+3+9+27+....+2187+6561 

=>A = 3⁰ + 3¹ + 3² + 3³ + .. . +3⁷ + 3⁸ 

 3A = 3¹ + 3² + 3³  + ... + 3⁸ + 3⁹

3A - A = (3¹ + 3² + 3³  + ... + 3⁸ + 3⁹)-(3⁰ + 3¹ + 3² + 3³ + .. . +3⁷ + 3⁸ )

2A = 3⁹ - 1 

A=\(\frac{3^9-1}{2}=\frac{19682}{2}=9841\)

1+3+9+27+....+2187+6561

Câu hỏi của nguyenphucthang - Toán lớp 4 - Học toán với OnlineMath

\(3A=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\)

\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{6561}\right)\)

\(2A=\dfrac{6560}{6561}\)

\(A=\dfrac{3280}{6561}\)

M = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)

=> 3M = \(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\)

=> 3M - M = ( \(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\)  ) - ( \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\))

2M = 1 - \(\frac{1}{6561}\)

2M = \(\frac{6560}{6561}\)

=> M = \(\frac{3280}{6561}\)

\(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+.......+\frac{1}{6561}\)

\(\Rightarrow M=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\)

\(\Rightarrow3M=3\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\right)\)

\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+............+\frac{1}{3^7}\)

\(\Rightarrow3M-M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..........+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-.......-\frac{1}{3^8}\)

\(\Rightarrow2M=1-\frac{1}{3^8}\)

\(\Rightarrow M=\frac{1-\frac{1}{3^8}}{2}\)

Vậy M = \(\frac{1-\frac{1}{3^8}}{2}\)

\(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{6561}\)

\(\Rightarrow3M=3\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{6561}\right)\)

\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)

\(\Rightarrow3M-M=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{6561}\right)\)

\(\Rightarrow2M=1-\frac{1}{6561}\)

\(\Rightarrow2M=\frac{6560}{6561}\)

\(\Rightarrow M=\frac{3280}{6561}\)

\(A=\dfrac{3}{5.6}+\dfrac{3}{6.7}+...+\dfrac{3}{91.92}\)

\(\Rightarrow A=3\left(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{91.92}\right)\)

\(\Rightarrow A=3\left(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{91}-\dfrac{1}{92}\right)\)

\(\Rightarrow A=3\left(\dfrac{1}{5}-\dfrac{1}{92}\right)\)

\(\Rightarrow A=3.\dfrac{87}{460}=\dfrac{261}{460}\)