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\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
=\(\frac{1.2.3...99}{2.3.4...100}\)
=\(\frac{1}{100}\)
Chúc bạn học giỏi nha
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)..........\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}.\frac{4}{3}.........\frac{100}{99}\)
\(=\frac{100}{2}=50\)
\(B=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).........\left(\frac{1}{100}-1\right)\)
\(=-\frac{1}{2}.-\frac{2}{3}..........-\frac{99}{100}\)
\(=\frac{-1}{100}\)
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)......\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(=\frac{3.4.5.....100}{2.3.4.....99}\)
\(=\frac{100}{2}=50\)
a)
\(=\frac{3}{2}.\frac{4}{3}......\frac{100}{99}=\frac{100}{2}=50\)
b)
\(=\frac{\left(-1\right)}{2}.\frac{\left(-2\right)}{3}.....\frac{\left(-99\right)}{100}=\frac{-1}{100}\)
\(\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{10}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)\(=\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\frac{1}{100}\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)\(=\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot0\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)=0
\(\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right).\left(\frac{1}{100}-\left(\frac{1}{2}\right)^2\right)......\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right)....\left(\frac{1}{100}-\left(\frac{1}{10}\right)^2\right)...\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right)...\left(\frac{1}{100}-\frac{1}{100}\right)...\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right).....0......\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=0\)
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow2B=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow2B-B=1-\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow B=1-\frac{1}{2^{100}}\)
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}\)
\(2B=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{99}\)
\(2B-B=\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{99}\right]-\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{100}\right]\)
\(B=1-\left(\frac{1}{2}\right)^{100}\)
\(B=1-\frac{1}{2^{100}}\)
Tham khảo nhé~