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Ta có: \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
Đặt A=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\)
2A=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{8.9.10}\)
2A=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}\) \(+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
2A=\(\frac{1}{1.2}-\frac{1}{9.10}\)
2A=\(\frac{22}{45}\)
A=\(\frac{22}{45}\div2\)
A=\(\frac{11}{45}\)
\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}\div\frac{11}{45}=\frac{23}{11}\)
Vậy x=\(\frac{23}{11}\)
Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100
4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4
4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)
4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100
4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101
4A=98.99.100.101
=>A=98.99.100.101/4
Rút gọn mỗi số hãng của số ta được :
\(C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
Vậy C = 100/101
\(C=\frac{4}{1.2.3}+\frac{8}{3.4.5}+\frac{12}{5.6.7}+...+\frac{200}{99.100.101}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{101}{101}-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(M=1.2.3+2.3.4+3.4.5+...+47.48.49\)
\(4M=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+47.48.49.\left(50-46\right)\)
\(=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+47.48.49.50-46.47.48.49\)
\(=47.48.49.50\)
\(M=\frac{47.48.49.50}{4}=1381800\)
( 10n ) chia het cho ( 5n - 3 )
=> ( 5n + 5n ) chia het cho ( 5n - 3 )
=> ( 5n - 3 + 5n - 3 + 6 ) chia het cho ( 5n - 3 )
=> [ 2.(5n-3) + 6 ] chia het cho ( 5n - 3 )
Ma (5n-3) chia het cho (5n - 3 )
=> 2(5n-3) chia het cho (5n-3)
=> 6 chia het cho (5n-3)
=> 5n - 3 thuoc U(6)
=> 5n - 3 thuoc { 1; 2;3;6 }
=> 5n thuoc { 0; 3 }
=> n = 0
Vay n = 0
P/s tham khao nha
I don't now
mik ko biết
sorry
......................
b,\(B=2^2+4^2+...+20^2\)
\(\Rightarrow B=2^2\left(1^2+2^2+...+10^2\right)\)
\(\Rightarrow B=4.\left[1.\left(2-1\right)+2.\left(3-1\right)+...+10.\left(11-1\right)\right]\)
\(\Rightarrow B=4\left(1.2-1+2.3-2+...+10.11-10\right)\)
\(\Rightarrow B=4\left[\left(1.2+2.3+...+10.11\right)-\left(1+2+...+10\right)\right]\)
\(\Rightarrow B=4\left(\frac{10.11.12}{3}-\frac{11.10}{2}\right)\)
\(A=1.2.3+3.4.5+5.6.7+...+99.100.+101\)
\(A=1.3\left(5-3\right)+3.5\left(7-3\right)+5.7\left(9-3\right)+...+99.100\left(103-3\right)\)
\(=\left(1.3.5+3.5.7+5.7.9+99.101.103\right)-\left(1.3.3+3.5.3+99.101.3\right)\)
\(=\left(15+99.101.103.105\right):8-3.\left(1.3+3.5+5.7+99.101\right)\)
\(=13517400-3.171650\)
\(=13002450\)
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