a)Rút gọn biểu thức A=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)
b) So sánh : A=\(\frac{20^{10}+1}{20^{10}-1}\) và B=\(\frac{20^{10}-1}{20^{10}-3}\)
Rút gọn:
A =\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)
B = \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
So sánh :
\(A=\frac{20^{10}+1}{20^{10}-1}vàB=\frac{20^{10}-1}{20^{10}-3}\)
Bài 1. Tính giá trị của biểu thức sau:
A=\(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
Bài 2. So sánh
A=\(\frac{20^{10}+1}{20^{10}-1}\)
B=\(\frac{20^{10}-1}{20^{10}-3}\)
Bài 3.Tính nhanh
P=\(\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}\)
Bài 1:
\(A=\frac{3333}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{3333}{101}\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{3333}{101}.\frac{4}{21}=\frac{1111.4}{101.7}=\frac{4444}{707}\)
Bài 2
\(A=\frac{2^{10}+1}{2^{10}-1}=\frac{2^{10}-1+2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(B=\frac{2^{10}-1}{2^{10}-3}=\frac{2^{10}-3+4}{2^{10}-3}=1+\frac{4}{2^{10}-3}\)
Ta thấy \(2^{10}-1>2^{10}-3\Rightarrow\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}< \frac{4}{2^{10}-3}\)
Từ đó \(\Rightarrow1+\frac{2}{2^{10}-1}< 1+\frac{4}{2^{10}-3}\Rightarrow A< B\)
Bài 3\(P=\frac{\left(\frac{2}{3}-\frac{1}{4}\right)+\frac{5}{11}}{\frac{5}{12}+\left(1-\frac{7}{11}\right)}=\frac{\frac{5}{12}+\frac{5}{11}}{\frac{5}{12}+\frac{4}{11}}=\frac{\frac{55+60}{11.12}}{\frac{55+48}{12.11}}=\frac{115}{103}\)
Bài 1 :So sánh
M=\(\frac{1}{2}\)+ \(\frac{1}{6}\)+..........+\(\frac{1}{2450}\) với 1
Bài 2:Rút gọn
1+\(\frac{1}{2}\)+\(\frac{1}{2^2}\)+ ........+ \(\frac{1}{2^{2012}}\)
Bài 3 : So sánh
A=\(\frac{20^{10}+1}{20^{10}-1}\)và B=\(\frac{20^{10}-1}{20^{10}-3}\)
Bài 4 :
Tìm n thuộc Z để \(\frac{3n+1}{3n+1}\)là số nguyên
Bài 5:
Chứng minh rằng phân số sau \(\frac{2n+1}{3n+1}\)là phân số tối giản với n thuộc Z
1.\(A=\frac{a}{b+c}=\frac{c}{a+b}=\frac{b}{c+a}\)
2. \(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
3. Hãy so sánh A và B
\(A=\frac{10^{2006}+1}{10^{2007}+1}\) \(B=\frac{10^{2007}+1}{10^{2008}+2}\)
So sánh: \(A=\frac{2^{10}+1}{20^{10}-1}\) và \(B=\frac{20^{10}-1}{20^{10}-3}\)
Ta có a/b >1 => a/b > a+n/b+n(a, b,n $\in$∈ N*)
B = 2010-1/2010-3 > 1 nên B = 2010-1/2010-3 > 2010-1+2/2010-3+2
= 2010+1/ 2010-1 = A
Vay \(A=\frac{2^{10}+1}{20^{10}-1}<\frac{20^{10}-1}{20^{10}-3}\)
So sánh A và B:
\(A=\frac{20^{10}+1}{20^{10}-1};B=\frac{20^{10}-1}{20^{10}-3}\)
So sánh A và B:
a,A=\(\frac{10^{2004}+1}{10^{2005}+1}\)
B=\(\frac{10^{2005}+1}{10^{2006}+1}\)
b,A=\(\frac{20^{10}+1}{20^{10}-1}\)
B=\(\frac{20^{10}-1}{20^{10}-3}\)
a) Ta có : 10A = \(\frac{10\left(10^{2004}+1\right)}{10^{2005}+1}=\frac{10^{2005}+10}{10^{2005}+1}=1+\frac{9}{10^{2005}+1}\)
Lại có 10B = \(\frac{10\left(10^{2005}+1\right)}{10^{2006}+1}=\frac{10^{2006}+10}{10^{2006}+1}=1+\frac{9}{10^{2006}+1}\)
Vì \(\frac{9}{10^{2005}+1}>\frac{9}{10^{2006}+1}\Rightarrow1+\frac{9}{10^{2005}+1}>1+\frac{9}{10^{2006}+1}\)
=> 10A > 10B
=> A > B
b) Ta có A = \(\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
Lại có B = \(\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1-\frac{2}{20^{10}-3}\)
=> A < B
Cảm ơn bạn rất nhiều nha
So sánh : \(A =\frac{20^{10} +1}{20^{10}-1} ; B =\frac{20^{10} -1}{20^{10} -3}\)
\(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
\(20^{10}-1>20^{10}-3\Rightarrow\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\)
=> A < B
So sánh A=\(\frac{20^{10}+1}{20^{10}-1}\)và B=\(\frac{20^{10}-1}{20^{10}-3}\)
ta thấy B>1 nên B=\(\frac{20^{10}-1}{20^{10}-3}\)>\(\frac{20^{10}-1+2}{20^{100}-3+2}\)=\(\frac{20^{10}+1}{20^{10}-1}\)=A
vậy B>A
nếu ko hiểu thì tham khảo trong SBT lớp 6 bài so sánh PS ấy