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Ta đặt: A = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
\(\Rightarrow3A=3+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(\Rightarrow3A-A=\left(3+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)\)
\(\Rightarrow2A=3-\frac{1}{3^4}\)
\(\Rightarrow A=\left(3-\frac{1}{3^4}\right):2\)
Giải
1+ 1 /3+1/9+1/27+1/81+1/243+1/729.
Đặt:
S = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
Nhân S với 3 ta có:
S x 3 = 3 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
Vậy:
S x 3 - S = 3 - 1/243
2S = 2186 / 729
S = 2186 / 729 : 2
S = 1093/729
A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
A * 3= 3* ( 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
A* 3 = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
A * 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
A * 2 = 1 - 1/ 729
A * 2 = 1/728
A = 1/728 : 2
A = 2/728
Nếu không quy đồng Mẫu thì ta quy đồng Tử
P/S: 2/728 VÀ 1/2
1/2 = 1*2/ 2*2
= 2/4
So sánh 2/4 và 2/278 ta thấy phân số 2/4 lớn hơn.
Vậy 1/2 > A
Đ/S: A = 2/728
1/2 > A
\(A=\frac{1}{3}+\frac{1}{3x3}+\frac{1}{3x3x3}+\frac{1}{3x3x3x3}+\frac{1}{3x3x3x3x3}+\frac{1}{3x3x3x3x3x3}.\)
\(3xA=1+\frac{1}{3}+\frac{1}{3x3}+\frac{1}{3x3x3}+\frac{1}{3x3x3x3}+\frac{1}{3x3x3x3x3}\)
\(2xA=3xA-A=1-\frac{1}{3x3x3x3x3x3}\)
\(A=\frac{1}{2}-\frac{1}{3x3x3x3x3x3}< \frac{1}{2}\)
A= 1+\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27} +\frac{1}{81}\)
=1+\(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
=> 3A-A=(\(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3}+1\))-(1+\(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\))
=>2A=3-\(\frac{1}{3^4}\)
=> A=(3-\(\frac{1}{3^4}\)):2
Bạn kiểm tra lại đề hộ. Nếu có phân số \(\frac{1}{4}\)thì chịu còn không có thì dễ.
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2048}\)
\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+...+\left(\frac{1}{1024}-\frac{1}{2048}\right)\)
\(A=1-\frac{1}{2048}\)
\(\Rightarrow\)\(A=\frac{2047}{2048}\)
\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3B-B=1-\frac{1}{2187}\)
\(2B=\frac{2186}{2187}\)
\(\Rightarrow B=\frac{2186}{4374}=\frac{1093}{2187}\)
bài 1 tính nhanh
mik xin sửa đề câu a thành thế này ~
\(a,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(A\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(A\cdot2-A=\) ( \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\) ) - ( \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\) )
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
\(b,\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
đặt \(B=\) \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(B\cdot3=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(B\cdot3-B=\) ( \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)) - \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\) )
\(B\cdot2=\) \(1-\frac{1}{729}\)
\(B\cdot2=\frac{728}{729}\)
\(B=\frac{728}{729}:2\)
\(B=\frac{364}{729}\)
\(c,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
ĐẶT \(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(C=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(C=\frac{1}{1}-\frac{1}{6}\)
\(C=\frac{5}{6}\)
{(1999x2001-1)/(1998+1999x2000)}x7/5
={[(1999x(2000+1)-1]/(1998+1999x2000)}...
={(1999x2000+1999-1)/(1998+1999x2000)}...
={(1999x2000+1998)/(1998+1999x2000)}x7...
=1x7/5
=7/5
Ta có: \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\right)\)
\(\Rightarrow2B=1-\frac{1}{3^6}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^6}}{2}\)
a)
\(\frac{32+16+8+4+2+1+128}{64}\)
\(\frac{191}{64}\)
B)
\(\frac{81+27+9+3+1+243}{243}\)
\(\frac{364}{243}\)
Mình lười làm qua :(