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\(A=\frac{1}{2}:\frac{4}{3}:\frac{-5}{4}:\frac{6}{5}:...:\frac{-101}{100}\)
<=> \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{-4}{5}\cdot\frac{5}{6}\cdot...\cdot\frac{-100}{101}\)
Trong biểu thức A có số số âm là (100-4):2 + 1 =49 số
Vậy A là số âm => \(A=-\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{100}{101}\right)\)
=> \(A=-\left(\frac{1}{2}\cdot\frac{3}{101}\right)=\frac{-3}{202}\)
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 \(\left(2\right)\) \(\Rightarrow A:B=\frac{2^{12}-1}{15}:\left(2^{12}-1\right)=\frac{1}{15}\)
Xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Khi đó:
\(1-\frac{2}{2.3}=\frac{1.4}{2.3}\) ; \(1-\frac{2}{3.4}=\frac{2.5}{3.4}\) ; ... ; \(1-\frac{2}{101.102}=\frac{100.103}{101.102}\)
\(\Rightarrow M=\frac{1.4}{2.3}\cdot\frac{2.5}{3.4}\cdot\cdot\cdot\frac{100.103}{101.102}\)
\(M=\frac{\left(1.2...100\right).\left(4.5...103\right)}{\left(2.3...101\right).\left(3.4...102\right)}=\frac{103}{101.3}=\frac{103}{303}\)
Vậy \(M=\frac{103}{303}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
\(=\frac{1}{100}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{100}\right)\)
Đặt : \(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}\)
\(A=\frac{1.2.3.4.....99}{2.3.4.5.....100}\)
\(A=\frac{1}{100}\)
Vậy : \(A=\frac{1}{100}\)
D = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{100^2}-1.\right)\)
=>\(-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{100^2}.\right)\)
=>\(-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{100^2-1}{100^2}\)
=>\(-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{99.101}{100^2}\right)\)
=>\(-\left(\frac{1.2.3...99}{2.3.4....100}\right)\left(\frac{3.4.5....101}{2.3.4....100}\right)\)
=>\(-\left(\frac{1}{100}.\frac{101}{2}\right)\)
=>\(D=-\frac{101}{200}\)