Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
biết làm bài 1 thôi
\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\cdot\cdot\cdot\times\left(\frac{1}{999}+1\right)\)
= \(\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdot\cdot\cdot\times\frac{1000}{999}\)
lượt bỏ đi còn :
\(\frac{1000}{2}=500\)
\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+...+16\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{16}.16.17:2=1+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}=\frac{2+3+4+...+17}{2}=\frac{152}{2}=76\)
\(d=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).........\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{3}.\frac{9}{2.4}.............\frac{10000}{99.101}\)
\(=\frac{2.2}{3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}............\frac{100.100}{99.101}\)
\(=\frac{2.3.4..........100}{2.3.4............99}.\frac{2.3.4...........100}{3.4...........101}\)
\(=100.\frac{2}{101}\)\(=\frac{200}{101}\)
\(C=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{1994}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{1993}{1994}\)
\(=\frac{1\times2\times3\times...\times1993}{2\times3\times4\times...\times1994}\)
\(=\frac{1}{1994}\) (Giản ước còn lại như này)
Câu b: Đặt \(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)\cdot...\cdot\left(\frac{1}{2004}-1\right)\)
Ta có: \(\frac{1}{2}-1=\left(-\frac{1}{2}\right);\frac{1}{3}-1=\left(-\frac{2}{3}\right);...;\frac{1}{2004}-1=\left(-\frac{2003}{2004}\right)\)
\(\Rightarrow B=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot...\cdot\left(-\frac{2003}{2004}\right)\)
Vì B là 2003 thừa số âm nhân lại với nhau nên B là số âm
\(\Rightarrow B=-\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2003}{2004}\right)=-\frac{1}{2004}\)
Câu a: Đặt \(A=1+2^4+2^8;B=1+2+2^2+...+2^{11}\)
\(\Rightarrow16A=2^4+2^8+2^{12}\) \(\Rightarrow15A=2^{12}-1\) \(\Rightarrow A=\frac{2^{12}-1}{15}\) \(\left(1\right)\)
\(\Rightarrow2B=2+2^2+2^3+...+2^{12}\) \(\Rightarrow B=2^{12}-1\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow A:B=\frac{2^{12}-1}{15}:\left(2^{12}-1\right)=\frac{1}{15}\)
Ta có (1-1/2).(1-1/3^2).(1-1/4^2).....(1-1/10^2)
=(2^2-1/2^2).(3^2-1/3^2).....(10^2-1/10)
=(1.3/2^2).(2.4/3^2).....(9.11/10^2)
=11/20
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\cdot\cdot\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\cdot\cdot\cdot\frac{100}{99}\)
\(=\frac{100}{2}=50\)
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}\)
\(=\frac{3.4.5...100}{2.3.4...99}\)
\(=\frac{100}{2}\)
\(=50\)