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A = \(\dfrac{1}{1+2+3}\)+\(\dfrac{1}{1+2+3+4}\)+...+ \(\dfrac{1}{1+2+...+2004}\)+ \(\dfrac{2}{2025}\)
A = \(\dfrac{1}{\left(1+3\right).3:2}\)+\(\dfrac{1}{\left(4+1\right).4:2}\)+...+ \(\dfrac{1}{\left(2024+1\right).2024:2}\)+\(\dfrac{2}{2025}\)
A = \(\dfrac{2}{3.4}\)+\(\dfrac{2}{4.5}\)+...+\(\dfrac{2}{2024.2025}\)+ \(\dfrac{2}{2025}\)
A = 2.(\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{2024.2025}\)) + \(\dfrac{2}{2025}\)
A = 2.(\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+...+ \(\dfrac{1}{2024}\) - \(\dfrac{1}{2025}\)) + \(\dfrac{2}{2025}\)
A = 2.(\(\dfrac{1}{3}\) - \(\dfrac{1}{2025}\)) + \(\dfrac{2}{2025}\)
A = \(\dfrac{2}{3}\) - \(\dfrac{2}{2025}\) + \(\dfrac{2}{2025}\)
A = \(\dfrac{2}{3}\)
a; A = (32.5 - 160: 22) + 2024
A = (9.5 - 160 : 4) + 2024
A = (45 - 40) + 2024
A = 5 + 2024
A = 2029
b; B = (-360) - (-87) + 69 - 87
B = -360 + 87 + 69 - 87
B = - (360 - 69) + (87 - 87)
B = - 291 + 0
B = -291
c; C = 182.26 - 82.26 + 500
C = 26.(182 - 82) + 500
C = 26.100 + 500
C = 2600 + 500
C = 3100
a, \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.100}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A< 2-\frac{1}{50}\)
\(A< 2\)
b, \(B=2+2^2+2^3+...+2^{30}\)
Ta có :\(B=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{29}+2^{30}\right)\)
\(B=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{29}\left(1+2\right)\)
\(B=2.3+2^3.3+...+2^{29}.3\)
\(B=3\left(2+2^3+...+2^{29}\right)\)chia hết cho 3(1)
Lại có\(B=\left(2+2^2+2^4\right)+...+\left(2^{28}+2^{29}+2^{30}\right)\)
\(B=2\left(1+2+4\right)+...+2^{28}\left(1+2+4\right)\)
\(B=2.7+...+2^{28}.7\)
\(B=7\left(2+...+2^{29}\right)\) chia hết cho 7 (2)
Mà (3,7)=1 (3)
Từ (1)(2)(3) => B chia hết cho 21
1+1/2.(1+2)+1/3.(1+2+3)+1/4.(1+2+3+4)+...+1/2023.(1+2+3+...+2023)
=1+1/2.(1+2).2/2+1/3.(1+3).3/2+1/4.(1+4).4/2+...+1/2023.(1+2+3+...+2023).2023/2
=2/2+3/2+4/2+...+2023/2
=2+3+4+...+2023/2
=2025.2022/2/2
=1023637,5
a) \(1,25^2-0,5^2+3,25^3\)
\(=\left(\frac{5}{4}\right)^2-\left(\frac{1}{2}\right)^2+\left(\frac{13}{4}\right)^3\)
\(=\frac{25}{16}-\frac{1}{4}+\frac{2197}{64}\)
\(=\frac{2281}{64}\)
b) \(1,2+2,4^2-\left(3\frac{1}{2}\right)^2\)
\(=\frac{6}{5}+\left(\frac{12}{5}\right)^2-\left(\frac{7}{2}\right)^2\)
\(=\frac{6}{5}+\frac{144}{25}-\frac{49}{4}\)
\(=\frac{-529}{100}\)