Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b, \(3737.43-4343.37=\left(37.101\right).43-\left(43.101\right).37=0\)
suy ra B = 0
c, \(D=\frac{2^{12}\left(13+65\right)}{2^{10}.104}+\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{2^{12}.78}{2^{10}.104}+\frac{3^{10}.16}{3^9.2^4}\)
\(=\frac{2^{12}.2.39}{2^{10}.2^3.13}+\frac{3^{10}.2^4}{3^9.2^4}=\frac{39}{13}+3=6\)
\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(=\frac{\frac{101.102}{2}}{51}\)
\(=101\)
Ta có \(\frac{2^{100}.13+65.2^{100}}{2^{98}.104}\)
\(=\frac{2^{100}.\left(13+65\right)}{2^{98}.2.52}\)
\(=\frac{2^{100}.78}{2^{99}.52}\)
\(=\frac{2.78}{52}\)
\(=3\)
\(\frac{2^{100}.13+65.2^{100}}{2^{98}.104}=\frac{2^{100}.\left(13+65\right)}{2^{98}.104}\)
\(=\frac{2^{100}.78}{2^{98}.104}\) \(=\frac{2^2.78}{104}=\frac{4.78}{104}\)
\(=\frac{78}{26}=3\)
\(T=\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right).......\left(\frac{1}{98}+1\right).\left(\frac{1}{99}+1\right)\)
\(T=\left(\frac{1}{2}+\frac{2}{2}\right).\left(\frac{1}{3}+\frac{3}{3}\right).\left(\frac{1}{4}+\frac{4}{4}\right).....\left(\frac{1}{98}+\frac{98}{98}\right).\left(\frac{1}{99}+\frac{99}{99}\right)\)
\(T=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{99}{98}.\frac{100}{99}\)
\(T=\frac{3.4.5....99.100}{2.3.4.....98.99}\)
\(T=\frac{100}{2}\)
\(T=50\)
Vậy T = 50
Chúc bạn học tốt!
\(A=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{98}-\frac{1}{100}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(A=2.\frac{49}{100}\)
\(A=\frac{49}{50}\)
A = 1/2 - 1/4 + 1/4 - 1/6 + ... + 1/98 - 1/100
A = 1/2 - 1/100
A = 49/100
Ta có: 1+1/2 +1/3 +...+1/98
=(1+1/98 )+(1/2 +1/97 )+(1/3 +1/96 )+...+(1/49 +1/50 )
=99/1.98 +99/2.97 +99/3.96 +...+99/49.50
=99(1/1.98 +1/2.97 +1/3.96 +...+1/49.50 )
⇒A=(1+1/2 +1/3 +...+1/98 ).2.3.4....98
=99(1/1.98 +1/2.97 +1/3.96 +...+1/49.50 ).2.3.4....98chia hết cho 99 (đpcm)
\(\frac{2^{100}.13+4^{50}.65}{2^{98}.104}=\frac{2^{100}.13+\left(2^2\right)^{50}.65}{2^{98}.104}=\frac{2^{100}.13+2^{100}.65}{2^{98}.104}=\frac{2^{100}.\left(13+65\right)}{2^{98}.104}=\frac{2^{100}.78}{2^{98}.104}=3\)