Tính A=\(\left(1+\frac{1}{1+2}\right).\left(1+\frac{1}{1+2+3}\right).\left(1+\frac{1}{1+2+3+4}\right)......\)\(+\left(1+\frac{1}{1+2+3+....997+998}\right)\)
3. Tính:
a) \(2^3-\left(\frac{1}{3}\right)^0.\sqrt{81}\)
b) \(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)...\left(\frac{1}{998}-1\right).\left(\frac{1}{999}-1\right)\)
a) 2^3-(1/3)^0.9
=8-(1/3)^0
=8-1
=7
b) mk quên cách giải rồi
sorry mai nha
Tính:
\(\left(\frac{1000}{1}+\frac{999}{2}+\frac{998}{3}+\frac{997}{4}+...+\frac{2}{999}+\frac{1}{1000}\right)\)\(:\)\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{1000}\right)\)
a)Tính: \({\left( {\frac{{ - 1}}{2}} \right)^5};{\left( {\frac{{ - 2}}{3}} \right)^4};{\left( { - 2\frac{1}{4}} \right)^3};{\left( { - 0,3} \right)^5};{\left( { - 25,7} \right)^0}\).
b)Tính: \({\left( { - \frac{1}{3}} \right)^2};{\left( { - \frac{1}{3}} \right)^3};{\left( { - \frac{1}{3}} \right)^4};{\left( { - \frac{1}{3}} \right)^5}\).
Hãy rút ra nhận xét về dấu của luỹ thừa với số mũ chẵn và luỹ thừa với số mũ lẻ của một số hữu tỉ âm.
a)
\(\begin{array}{l}{\left( {\frac{{ - 1}}{2}} \right)^5} = \frac{{{{\left( { - 1} \right)}^5}}}{{{2^5}}} = \frac{{ - 1}}{{32}};\\{\left( {\frac{{ - 2}}{3}} \right)^4} = \frac{{{{\left( { - 2} \right)}^4}}}{{{3^4}}} = \frac{{16}}{{81}};\\{\left( { - 2\frac{1}{4}} \right)^3} = {\left( {\frac{{ - 9}}{4}} \right)^3} = \frac{{{{\left( { - 9} \right)}^3}}}{{{4^3}}} = \frac{{-729}}{{64}};\\{\left( { - 0,3} \right)^5} = {\left( {\frac{{ - 3}}{{10}}} \right)^5} = \frac{{ - 243}}{{100000}};\\{\left( { - 25,7} \right)^0} = 1\end{array}\)
b)
\(\begin{array}{l}{\left( { - \frac{1}{3}} \right)^2} = \frac{1}{9};\\{\left( { - \frac{1}{3}} \right)^3} = \frac{{ - 1}}{{27}};\\{\left( { - \frac{1}{3}} \right)^4} = \frac{1}{{81}};\\{\left( { - \frac{1}{3}} \right)^5} = \frac{{ - 1}}{{243}}.\end{array}\)
Nhận xét:
+ Luỹ thừa của một số hữu tỉ âm với số mũ chẵn là một số hữu tỉ dương.
+ Luỹ thừa của một số hữu tỉ âm với số mũ lẻ là một số hữu tỉ âm.
Thực hiện phép tính :
a, A =\(\left(1:\frac{5^2}{10^2}\right).\left(1\frac{1}{1}\right)^2+25.\left[1:\left(\frac{4}{3}\right)^2:\left(\frac{5}{4}\right)^3\right]:\left(1:\frac{-8}{27}\right)\)
b, B =\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{100^2}\right)\)
a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)
\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)
\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
Tính hợp lí:
\(\left(1+\frac{1}{1+2}\right)\cdot\left(1+\frac{1}{1+2+3}\right)\cdot...\cdot\left(1+\frac{1}{1+2+3+...+997}\right)\)
\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).......\left(\frac{1}{998}-1\right).\left(\frac{1}{999}-1\right)\)
Lời giải:
Ta có:
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{998}-1\right)\left(\frac{1}{999}-1\right)=\frac{1-2}{2}.\frac{1-3}{3}.....\frac{1-998}{998}.\frac{1-999}{999}\)
\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-997}{998}.\frac{-998}{999}\)
\(=\frac{(-1)(-2)(-3)....(-998)}{2.3.4...999}=\frac{1.2.3....998}{2.3.4...999}=\frac{1}{999}\)
tính các tích sau với nEN, n lớn hơn bằng 2
a)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n}\right)\)
b)\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1-\frac{1}{n}\right)\)
c)\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
TRÌNH BÀY GIÚP MÌNH NHA