1+1/3+1/9+1/27+1/81+1/243+1/729
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\(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}\)+\(\dfrac{1}{243}+\dfrac{1}{729}\)=\(\dfrac{1093}{729}\)
\(=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
\(\dfrac{729}{729}+\dfrac{243}{729}+\dfrac{81}{729}+\dfrac{27}{729}+\dfrac{9}{729}+\dfrac{3}{729}+\dfrac{1}{729}\)
\(=\dfrac{\left(729+243+81+27+9+3+1\right)}{729}=\dfrac{1084}{729}\)
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\) + \(\dfrac{1}{729}\)
3 \(\times\) A = 3 + 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\)
3 \(\times\) A - A = 3 - \(\dfrac{1}{729}\)
A \(\times\)(3-1) = \(\dfrac{2186}{729}\)
A \(\times\) 2 = \(\dfrac{2186}{729}\)
A = \(\dfrac{2186}{729}\): 2
A = \(\dfrac{1093}{729}\)
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(=\frac{243}{729}+\frac{81}{729}+\frac{27}{729}+\frac{3}{729}\)
\(=\frac{243+81+27+3}{729}=\frac{354}{729}\)
\(\frac{1}{3}\)+ \(\frac{1}{9}\)+ \(\frac{1}{27}\)+ \(\frac{1}{81}\)+ \(\frac{1}{243}\)+ \(\frac{1}{729}\)
= \(\frac{243}{729}\)+ \(\frac{81}{729}\)+ \(\frac{27}{729}\)+ \(\frac{9}{729}\)+ \(\frac{3}{729}\)+ \(\frac{1}{729}\)
= \(\frac{\left(243+27\right)+\left(81+9\right)+\left(3+1\right)}{729}\)
= \(\frac{270+90+4}{729}\)
=\(\frac{364}{729}\)
:)
1/3+1/9=3/9+1/9=4/9+1/27=12/27+1/27=13/27+181=39/81+1/81=40/81+1/243=120/243+1/243=121/243+1/729=363/729+1/729=364/729
Đặt S=\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
3S=\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
3S-S=\(1-\frac{1}{729}\)
2S=\(\frac{728}{729}\)
S=\(\frac{728}{729}:2\)
S=\(\frac{364}{729}\)
Câu trả lời hay nhất: Đặt A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
A x 3 = 3 x (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
A x 3 - A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - 1/3 - 1/9 - 1/27 - 1/81 - 1/243 - 1/729
= 1 - 1/729
A x 2 = 728/729
A = 364/729
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(=\frac{243}{729}+\frac{81}{729}+\frac{27}{729}+\frac{9}{729}+\frac{3}{729}+\frac{1}{729}\)
\(=\frac{4}{9}+\frac{27}{729}+\frac{9}{729}+\frac{3}{729}+\frac{1}{729}\)
\(=\frac{13}{27}+\frac{9}{729}+\frac{3}{729}+\frac{1}{729}\)
\(=\frac{40}{81}+\frac{3}{729}+\frac{1}{729}\)
\(=\frac{121}{243}+\frac{1}{729}\)
\(=\frac{364}{729}\).
Ủng hộ nha !
A= 1/3+1/9+1/27+1/81+1/243+1/729
A= (1-1/3)+(1/3-1/9)+(1/9-1/27)+(1/27-1/81)+(1/81-1/243)+(1/243-1/729)
A= 1-1/3+1/3-1/9+1/9-1/27+1/27-1/81+1/81-1/243+1/243-1/729
Triệt tiêu từ 1/3 đến 1/243
A= 1-1/729
A = 729/729 - 1/729= 91
Đ/S 9 mình nha
\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}\\ \Rightarrow A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\\ \Rightarrow\dfrac{1}{3}A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^7}\\ \Rightarrow\dfrac{1}{3}A-A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^7}-1-\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^6}\\ \Rightarrow-\dfrac{2}{3}A=\dfrac{1}{3^7}-1\\ \Rightarrow A=\left(\dfrac{1}{2187}-1\right):\left(-\dfrac{2}{3}\right)\\ \Rightarrow A=\left(-\dfrac{2186}{2187}\right):\left(-\dfrac{2}{3}\right)\\ \Rightarrow A=\dfrac{1093}{729}\)
Các bạn làm như vậy với các cháu học sinh lớp 4, 5 là ko làm đc. KQ tính bằng 1093/729 là đúng nhưng PP làm chưa đúng.
Mình hướng dẫn con mình làm như thế này là phù hợp với kiến thức lớp 4:
Ta tách phân số như sau:
= (5/3-2/3) + (2/3-1/3) + (1/3-2/9) + (2/9-5/27) + (5/27-14/81) + (14/81-41/243) + (41/243-122/729)
Sau khi rút gọn ta còn:
= 5/3 - 122/729
= (5*243-122)/729
= 1093/729