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a) \(\frac{\left(-1\right)}{4}^2+\frac{3}{8}.\left(\frac{-1}{6}\right)-\frac{3}{16}:\left(\frac{-1}{2}\right)=\left(\frac{-1}{4}\right)^2+\left(\frac{-3}{68}\right)-\left(\frac{-3}{8}\right)=\left(\frac{1}{16}\right)+\left(\frac{-3}{68}\right)-\left(\frac{-3}{8}\right)=\frac{5}{272}-\left(\frac{-3}{8}\right)=\frac{107}{272}\)
câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)
Lời giải:
Xét công thức tổng quát:
$1+2+3+...+n=\frac{n(n+1)}{2}$
$\Rightarrow 1-\frac{1}{1+2+3+...+n}=1-\frac{2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}$
Thay $n=2,3,...,2006$ ta thu được:
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{2005.2008}{2006.2007}\)
\(=\frac{(1.2.3...2005)(4.5.6...2008)}{(2.3.4...2006)(3.4.5...2007)}=\frac{1}{2006}.\frac{2008}{3}=\frac{1004}{3009}\)
Lời giải:
Xét công thức tổng quát:
$1+2+3+...+n=\frac{n(n+1)}{2}$
$\Rightarrow 1-\frac{1}{1+2+3+...+n}=1-\frac{2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}$
Thay $n=2,3,...,2006$ ta thu được:
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{2005.2008}{2006.2007}\)
\(=\frac{(1.2.3...2005)(4.5.6...2008)}{(2.3.4...2006)(3.4.5...2007)}=\frac{1}{2006}.\frac{2008}{3}=\frac{1004}{3009}\)
\(A=(1-\frac{1}{1+2})(1-\frac{1}{1+2+3})(1-\frac{1}{1+2+3+4})...(1-\frac{1}{1+2+3+...+2006})\)
\(A=(1-\frac{1}{3})(1-\frac{1}{6})(1-\frac{1}{10})...(1-\frac{1}{2013021})\)
\(A=\frac{2}{3}\cdot\frac{5}{6}\cdot\frac{9}{10}....\frac{2013020}{2013021}\)
Sorry bạn máy tính mình có chút vấn đề để mk làm tiếp :
\(A=\frac{4}{6}\cdot\frac{10}{12}\cdot\frac{18}{20}....\cdot\frac{4026040}{4026042}\)
\(A=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot\frac{3\cdot6}{4\cdot5}\cdot...\cdot\frac{2005\cdot2008}{2006\cdot2007}\)
\(A=\frac{1\cdot2\cdot3\cdot...\cdot2005}{2\cdot3\cdot4\cdot...\cdot2006}\cdot\frac{4\cdot5\cdot6\cdot...\cdot2008}{3\cdot4\cdot5\cdot...\cdot2007}\)
\(A=\frac{1}{2006}\cdot\frac{2008}{3}=\frac{1004}{3009}\)
P/S : Hoq chắc :>
Giải:
1) \(7^8.\left(-\dfrac{1}{7}\right)^8\)
\(=7^8.\left(\dfrac{1}{7}\right)^8\)
\(=7^8.\dfrac{1^8}{7^8}\)
\(=1\)
2) \(\left(\dfrac{4}{3}\right)^{10}.\left(-\dfrac{3}{4}\right)^{10}\)
\(=\left(\dfrac{4}{3}\right)^{10}.\left(\dfrac{3}{4}\right)^{10}\)
\(=\dfrac{4^{10}}{3^{10}}.\dfrac{3^{10}}{4^{10}}\)
\(=1\)
3) \(\left(-\dfrac{7}{2}\right)^{2006}.\left(-\dfrac{2}{7}\right)^{2006}\)
\(=\left(\dfrac{7}{2}\right)^{2006}.\left(\dfrac{2}{7}\right)^{2006}\)
\(=1\)
4) \(\left(-\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\left(\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\dfrac{5^{2007}.13^{2006}}{13^{2007}.5^{2006}}\)
\(=\dfrac{5}{13}\)
Vậy ...
Ta thấy: \(1-\frac{1}{1+2+3+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(\Rightarrow A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)
\(=\left(\frac{\left(2-1\right)\left(2+2\right)}{2\left(2+1\right)}\right)\left(\frac{\left(3-1\right)\left(3+2\right)}{3\left(3+1\right)}\right)\left(\frac{\left(4-1\right)\left(4+2\right)}{4\left(4+1\right)}\right)...\left(\frac{\left(2006-1\right)\left(2006+2\right)}{2006\left(2006+1\right)}\right)\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{2005.2008}{2006.2007}=\frac{\left(1.2.3...2005\right)\left(4.5.6...2008\right)}{\left(2.3.4...2006\right)\left(3.4.5...2007\right)}\)
\(=\frac{1.2008}{2006.3}=\frac{1004}{1003.3}=\frac{1004}{3009}\)
Vậy \(A=\frac{1004}{3009}\)