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\(A=5+5^2+5^3+5^4+5^5+5^6+5^7+5^8+5^9\)
\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+\left(5^7+5^8+5^9\right)\)
\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+5^7\left(1+5+5^2\right)\)
\(=5.31+5^4.31+5^7.31=31.\left(5+5^4+5^7\right)\)chia hết cho 31
Vậy A chia 31 dư 0
\(S=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+8}\)
\(=1+\frac{1}{\left(1+2\right).3.\frac{1}{2}}+\frac{1}{\left(1+3\right).3.\frac{1}{2}}+...+\frac{1}{\left(1+8\right).8.\frac{1}{2}}\)
\(=1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{8.9}\)
\(=1+2.\left(\frac{3-2}{2+3}+\frac{4-3}{3.4}+...+\frac{9-8}{8.9}\right)\)
\(=1+2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{9}\right)\)
\(=1+2.\frac{7}{18}=1+\frac{7}{9}=\frac{16}{9}\)
2/
S = 2 + 22 + 23 +...+ 299
= (2+22+23) +...+ (297+298+299)
= 2(1+2+22)+...+297(1+2+22)
= 2.7 +...+ 297.7
= 7(2+...+297) chia hết cho 7
S = 2+22+23+...+299
= (2+22+23+24+25)+...+(295+296+297+298+299)
= 2(1+2+22+23+24)+...+295(1+2+22+23+24)
= 2.31+...+295.31
= 31(2+...+295) chia hết cho 31
3/
A = 1+5+52+....+5100 (1)
5A = 5+52+53+...+5101 (2)
Lấy (2) - (1) ta được
4A = 5101 - 1
A = \(\frac{5^{101}-1}{4}\)
4/
Đặt A là tên của biểu thức trên
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
........
\(\frac{1}{8^2}< \frac{1}{7.8}=\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}< 1\)
Vậy...
5/
a, Gọi UCLN(n+1,2n+3) = d
Ta có : n+1 chia hết cho d => 2(n+1) chia hết cho d => 2n+2 chia hết cho d
2n+3 chia hết cho d
=> 2n+2 - (2n+3) chia hết cho d
=> -1 chia hết cho d => d = {-1;1}
Vậy...
b, Gọi UCLN(2n+3,4n+8) = d
Ta có: 2n+3 chia hết cho d => 2(2n+3) chia hết cho d => 4n+6 chia hết cho d
4n+8 chia hết cho d
=> 4n+6 - (4n+8) chia hết cho d
=> -2 chia hết cho d => d = {1;-1;2;-2}
Mà 2n+3 lẻ => d lẻ => d khác 2;-2 => d = {1;-1}
Vậy...
bài 1:
\(\frac{6}{11}+\frac{1}{3}+\frac{5}{11}\)
\(=\left(\frac{6}{11}+\frac{5}{11}\right)+\frac{1}{3}\)
\(=\frac{11}{11}+\frac{1}{3}=1+\frac{1}{3}=\frac{3}{3}+\frac{1}{3}=\frac{4}{3}\)
bài 2:
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)
\(=\left(\frac{1}{2}+\frac{1}{20}\right)+\left(\frac{1}{6}+\frac{1}{12}\right)\)
\(=\frac{11}{20}+\frac{1}{4}=\frac{11}{20}+\frac{5}{20}=\frac{15}{20}=\frac{3}{4}\)
bài 3:
a) \(\frac{3}{2}\cdot\frac{4}{5}\cdot\frac{2}{3}=\left(\frac{3}{2}\cdot\frac{2}{3}\right)\cdot\frac{4}{5}=1\cdot\frac{4}{5}=\frac{4}{5}\)
b) \(\frac{6}{7}\cdot\frac{5}{3}\cdot\frac{7}{6}=\left(\frac{6}{7}\cdot\frac{7}{6}\right)\cdot\frac{5}{3}=1\cdot\frac{5}{3}=\frac{5}{3}\)
bài 4:
a) \(\frac{2}{5}\cdot\frac{1}{4}+\frac{3}{4}\cdot\frac{2}{5}=\frac{2}{5}\cdot\left(\frac{1}{4}+\frac{3}{4}\right)=\frac{2}{5}\cdot1=\frac{2}{5}\)
b) \(\frac{6}{11}:\frac{2}{3}+\frac{5}{11}:\frac{2}{3}=\left(\frac{6}{11}+\frac{5}{11}\right):\frac{2}{3}=1:\frac{2}{3}=\frac{3}{2}\)
Bài 1:
6/11 + 1/3 + 5/11
= ( 6/11 + 5/11) + 1/3
= 11/11 + 1/3
= 1 + 1/3
= 3/3 +1/3
= 4/3
Bài 2:
1/2 + 1/6 + 1/12 + 1/20
= ( 1/2 + 1/6 + 1/12 ) + 1/20
= ( 6/12 + 2/12 + 1/12 ) + 1/20
=9/12 + 1/20
= 3/4 +1/20
= 15/20 + 1/20
= 16/20 = 4/5
Bài 3:
a) \(\frac{3}{2}\times\frac{4}{5}\times\frac{2}{3}\) \(=\left(\frac{3}{2}\times\frac{2}{3}\right)\times\frac{4}{5}\)\(=1\times\frac{4}{5}=\frac{4}{5}\)
b) \(\frac{6}{7}\times\left(\frac{5}{3}\times\frac{7}{6}\right)\) \(=\frac{6}{7}\times\frac{35}{18}\)\(=\frac{1\times5}{7\times3}=\frac{5}{21}\)
Bài 4:
a) 2/5 x 1/4 + 3/4 x 2/5
= 2/5 x ( 1/4 + 3/4)
= 2/5 x 1
= 2/5
b) 6/11 : 2/3 +5/11 : 2/3
= ( 6/11 + 5/11) x 3/2
= 11/11 x 3/2
= 1 x 3/2
= 3/2
....
