bai 1 tinh nhanh
1+\(\frac{1}{3}\)+\(\frac{1}{9}\)+\(\frac{1}{27}\)+\(\frac{1}{81}\)+\(\frac{1}{243}\)
ban nao giai duoc viet loi giai ra minh tich nhieu nha
\(A=1+\frac{1}{2}+\frac{1}{2^2} +\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
ghi loi giai day du nha may ban
ai ghi du loi giai va nhanh minh se tick(toan lop 6 nhe)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
k nha
Nhân 2A lên rồi lấy 2A-A là ra kết quả
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(\Rightarrow A=2A-A\)
\(\Rightarrow A=2-\frac{1}{2^{2012}}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+....+\frac{1}{2010}}\)
Cac ban giai giup minh voi minh minh kiem tr roi
giai co loi giai nha
minh kich dung cho
Bước1: Chứng minh: x>ln(1+x)>x-x^2/2 (khảo sát hàm lớp 12)
Bước2: Đặt A=1+1/2+1/3+...+1/N.
B=1+1/2^2+1/3^2+...+1/N^2.
C=1+1/1.2+1/2.3+...+1/(N-1).N
D=ln(1+1)+ln(1+1/2)+ln(1+1/3)+...
...+ln(1+1/N).
Bước 3: Nhận xét: 1/k(k+1)=1/k-1/(k+1)
suy ra C=2-1/N <2
Bước 4: Nhận xét ln(k+1)-lnk=ln(1+1/k)
suy ra D=ln(N+1)
Bước 5: Nhận xét B<C<2
Bước 6: Chứng minh A->+oo (Omerta_V đã CM)
Bước 7: Từ Bước1 suy ra:
A>D>A-1/2B>A-1.
Bước 8: Vậy A xấp sỉ D với sai số tuyệt đối bằng 1.
Mà A->+oo. Nên khi N rất lớn thì sai số tương đối có thể coi là 0.
Cụ thể hơn Khi N>2^k thì sai số tương đối < k/2
Vậy khi N lớn hơn 1000000 thì ta có thể coi A=ln(N+1).
vậy đáp án là 5
Tính nhanh:
\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
Giúp mk nha!
\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(=\frac{729}{729}+\frac{243}{729}+\frac{81}{729}+\frac{27}{729}+\frac{9}{729}+\frac{3}{729}+\frac{1}{729}\)
\(=\frac{729+243+81+27+9+3+1}{729}\)
\(=\frac{1093}{729}\)
gọi biểu thức trên là A
ta có : A = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\) (1)
\(\frac{1}{3}\)x A =\(\frac{1}{3}\)+\(\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\) (2)
lấy (1) - (2)
\(\frac{2}{3}xA\)= 1 - \(\frac{1}{2187}\)
\(\frac{2}{3}xA\)= \(\frac{2186}{2187}\)
A = \(\frac{2186}{2187}:\frac{2}{3}\)
A = \(\frac{1093}{729}\)
Đặt\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow3A=3+1+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3A-A=3-\frac{1}{729}\)
\(\Rightarrow2A=\frac{2186}{729}\)
\(\Rightarrow A=\frac{1093}{729}\)
Tính nhanh:
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\text{Đặt : }A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3A-A=1-\frac{1}{729}\)
\(\Rightarrow2A=\frac{728}{729}\)
\(\Rightarrow A=\frac{728}{729}:2=\frac{364}{729}\)
\(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Rightarrow2A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}\)
\(\Rightarrow2A-A=\frac{1}{3^1}-\frac{1}{3^7}\)
\(\Rightarrow A=\frac{1}{3^1}-\frac{1}{3^7}\)
Tính nhanh \(\frac{1}{1}+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)
Gọi tong trên là A
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}+\frac{1}{7129}+\frac{1}{2187}\)
\(3A=\frac{1}{3}+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{729}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}-\frac{1}{243}-\frac{1}{729}-\frac{1}{2187}\)
\(2A=1-\frac{1}{2187}\)
\(2A=\frac{2186}{2187}\)
\(A=\frac{2186}{2187}:2\)
\(A=\frac{1093}{2187}\)
Vậy tổng A = \(\frac{1093}{2187}\)
\(3y=3\cdot\frac{1}{1}+3\cdot\frac{1}{3}+3\cdot\frac{1}{9}+...+3\cdot\frac{1}{729}+3\cdot\frac{1}{2187}\)
\(=3+\frac{1}{1}+\frac{1}{3}...+\frac{1}{729}\)
=> \(3y-y=3+\frac{1}{1}+\frac{1}{3}+..+\frac{1}{729}-\frac{1}{1}-\frac{1}{3}-...-\frac{1}{2187}\)
<=> 2y = 3- 1/2187
=> y = \(\frac{3-\frac{1}{2187}}{2}\)
\(\text{Đ}\text{ặt} A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)
\(\Rightarrow2187A=2187+729+243+81+27+9+3+1\)
\(\Leftrightarrow2187A=3280\)
\(\Leftrightarrow A=\frac{3280}{2187}\)
Chắc chắn 100% luôn
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
Tính nhanh na
tổng các ps trên là ; \(\frac{364}{729}\)
đặt biểu thức đó là X
ta có :
\(3X=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3X-X=1-\frac{1}{729}\)
\(\Rightarrow X=\frac{728}{729}.\frac{1}{2}=\frac{364}{729}\)
Tính nhanh tổng sau \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
=\(1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
=\(\frac{3^6}{3^6}+\frac{3^5}{3^6}+\frac{3^4}{3^6}+\frac{3^3}{3^6}+\frac{3^2}{3^6}+\frac{3^1}{3^6}+\frac{3^0}{3^6}\)
=\(\frac{3^6+3^5+3^4+3^3+3^2+3+1}{3^6}\)
=\(\frac{729+243+81+27+9+3}{729}\)
=\(\frac{1093}{729}\)
nha.
1+1/3+1/9+1/27+1/81+1/243+1/729
=729/729+243/729+81/729+27/729+9/729+3/729+1/729
=1093/729
Cho A =\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+.......+\frac{1}{70}\)
CMR :
a , A > \(\frac{4}{3}\)
b , A < 2,5
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A = \(\frac{1}{3}\)+ \(\frac{1}{9}\)+ \(\frac{1}{27}\) + \(\frac{1}{81}\)+ ................ + \(\frac{1}{6561}\)
Cac ban giai ho minh nhe, minh can gap trong toi nay.
giai cac buoc ra nhe !!!
Thank you very much .
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+....+\frac{1}{6561}\)
\(\Rightarrow\)\(3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....+\frac{1}{2187}\)
\(\Rightarrow\)\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{2187}\right)-\left(\frac{1}{3}+\frac{1}{9}+....+\frac{1}{6561}\right)\)
\(\Rightarrow\)\(2A=1-\frac{1}{6561}=\frac{6560}{6561}\)
\(\Rightarrow\)\(A=\frac{3280}{6561}\)