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D = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{100^2}-1.\right)\)
=>\(-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{100^2}.\right)\)
=>\(-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{100^2-1}{100^2}\)
=>\(-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{99.101}{100^2}\right)\)
=>\(-\left(\frac{1.2.3...99}{2.3.4....100}\right)\left(\frac{3.4.5....101}{2.3.4....100}\right)\)
=>\(-\left(\frac{1}{100}.\frac{101}{2}\right)\)
=>\(D=-\frac{101}{200}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
\(=\frac{1}{100}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{100}\right)\)
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 : \(A=\frac{1}{100}\)
\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)
\(M=1-\frac{1}{7}=\frac{6}{7}\)
Mình làm câu 1 thoi nha!
1.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
a) -1 - 2 + 3 + 4 - 5 - 6 + 7 + 8 - 9 - 10 + 11 + 12 - ... - 2013 - 2014 + 2015 + 2016
= ( -1 - 2 + 3 + 4 ) - ( 5 + 6 - 7 - 8 ) - ( 9 + 10 - 11 - 12 ) - .......... - ( 2013 + 2014 - 2015 - 2016 )
= 4 - ( -4 ) - ( -4 ) - ......... - ( -4 )
= 4 + 4 + 4 +....... + 4
= { [ ( 2016 - 1 ) : 1 + 1 ] : 4 } . 4
= { [ 2015 : 1 + 1 ] : 4 } . 4
= { 2016 : 4 } . 4
= 504 . 4
= 2016
b) \(\left(\frac{1}{2}-1\right):\left(\frac{1}{3}-1\right):\left(\frac{1}{4}-1\right):\left(\frac{1}{5}-1\right):.........:\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}:\frac{-2}{3}:\frac{-3}{4}:\frac{-4}{5}:......:\frac{-99}{100}\)
\(=\frac{-1}{2}.\frac{3}{-2}.\frac{4}{-3}.\frac{5}{-4}.......\frac{100}{-99}\)
\(=\frac{-1.3.4........100}{2.2.3.4......99}\)
\(=\frac{-1.100}{2.2}\)
\(=\frac{-100}{4}\)
\(=-25\)
a) -1-2+3+4-5-6+7+8+...+2016=-3+3-7+7-...-2016+2016=0
b) \(\left(\frac{1}{2}-1\right):...:\left(\frac{1}{100}-1\right)=\frac{-1}{2}:\frac{-2}{3}:\frac{-3}{4}:...:\frac{-99}{100}\)
\(=\)\(\frac{-1}{2}.\frac{-3}{2}.....\frac{-100}{99}=\frac{-1}{2}.\left(-50\right)=25\)
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 \(\left(2\right)\) \(\Rightarrow A:B=\frac{2^{12}-1}{15}:\left(2^{12}-1\right)=\frac{1}{15}\)
\(A=\frac{1}{2}:\frac{4}{3}:\frac{-5}{4}:\frac{6}{5}:...:\frac{-101}{100}\)
<=> \(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{-4}{5}\cdot\frac{5}{6}\cdot...\cdot\frac{-100}{101}\)
Trong biểu thức A có số số âm là (100-4):2 + 1 =49 số
Vậy A là số âm => \(A=-\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{100}{101}\right)\)
=> \(A=-\left(\frac{1}{2}\cdot\frac{3}{101}\right)=\frac{-3}{202}\)
thanks bn nhiều nha Hiếu