1093/729

\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

\(3\times A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(3\times A-A=3-\frac{1}{729}=\frac{2186}{729}\)

\(2\times A=\frac{2186}{729}=>A=\frac{1093}{729}\)

Mình nghĩ là đề sai. Các phân số đều có mẫu là lũy thừa của 3 vì thế 1/143 phải là 1/243 chứ

= ( 1 +729 + 1/143 ) + ( 1/3 + 1/9 + 1/27 + 1/81 )

= ( 730 + 1/143 ) + ( 27/81 + 9/81 + 3/81 + 1/81 )

= ( 730 + 1/143 ) + 40/81

= 104391/143 + 40/81

= 730, 5008202

\(x\) \(\times\) \(\dfrac{1}{4}\) = 6 : 1 : 2

\(x\) \(\times\) \(\dfrac{1}{4}\) = 6:2

\(x\) \(\times\) \(\dfrac{1}{4}\) =  3

\(x\)         = 3 : \(\dfrac{1}{4}\)

\(x\)        = 12

Mình nhầm đoạn 6:1/2.nhờ bạn giải lại hộ mình với

\(x\)\(\times\) \(\dfrac{1}{4}\) = 6 : \(\dfrac{1}{2}\)

\(x\) \(\times\) \(\dfrac{1}{4}\) = 6 \(\times\) \(\dfrac{2}{1}\)

\(x\) \(\times\) \(\dfrac{1}{4}\) = 12

\(x\)       =  12 : \(\dfrac{1}{4}\)

\(x\)      = 48

Tính nhanh:

                         A =         \(\dfrac{2}{3}\) + \(\dfrac{2}{9}\) + \(\dfrac{2}{27}\) + \(\dfrac{2}{81}\) + \(\dfrac{2}{243}\) + \(\dfrac{2}{729}\)

                  A \(\times\) 3 =  2  +  \(\dfrac{2}{3}\) + \(\dfrac{2}{9}\) + \(\dfrac{2}{27}\) + \(\dfrac{2}{81}\) + \(\dfrac{2}{243}\) 

            A \(\times\) 3 - A = 2 - \(\dfrac{2}{729}\)

           A \(\times\) ( 3 - 1) = 2 \(\times\) ( 1 - \(\dfrac{1}{729}\))

             A \(\times\) 2       = 2 \(\times\) \(\dfrac{728}{729}\)

            A = 2 \(\times\) \(\dfrac{728}{729}\) : 2

           A = \(\dfrac{728}{729}\)

c, tính:

 \(\dfrac{4}{5}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)

= \(\dfrac{4}{5}\) + \(\dfrac{1}{3}\)

= \(\dfrac{17}{15}\)

2 điểm!?

thi hay sao?

S= 1/1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

S= 3 x ( 1/1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 )

S = 3 +1 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

S= 3 +1 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 - 1 - 1/9 -1/27 - 1/81 - 1/243 - 1/729

S = 3 - 1/729 

S= 142/729

\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

\(=\frac{729}{729}+\frac{243}{729}+\frac{81}{729}+\frac{27}{729}+\frac{9}{729}+\frac{3}{729}+\frac{1}{729}\)

\(=\frac{729+243+81+27+9+3+1}{729}\)

\(=\frac{1087}{729}\)

Đặt \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

\(3A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(3A-A=3-\frac{1}{729}\)nên \(2A=3-\frac{1}{729}\)

kHI ĐÓ \(A=\left(3-\frac{1}{729}\right):2=\frac{3}{2}-\frac{1}{1458}=\frac{2197}{1458}-\frac{1}{1458}=\frac{2196}{1458}\)

 \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\)\(\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

\(=\frac{729}{729}+\frac{243}{729}+\)\(\frac{81}{729}+\frac{27}{729}+\frac{9}{729}+\)\(\frac{3}{729}\)\(+\frac{1}{729}\)

\(=\frac{1093}{729}\)