Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha
1, 7A = 7+7^2+7^3+....+7^2008
6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1
=> A = (7^2008-1)/6
Tk mk nha
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)
\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)
\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)
\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)
\(=-\dfrac{38}{42}\)
\(=-\dfrac{19}{21}\)
\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)
\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)
\(=\dfrac{13}{13}+11\)
\(=1+11\)
\(=12\)
\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)
\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)
\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)
\(=\dfrac{28}{7}-\dfrac{31}{9}\)
\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)
\(=\dfrac{252}{63}-\dfrac{217}{63}\)
\(=\dfrac{35}{63}\)
\(=\dfrac{5}{9}\)
\(5,\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2}{3}-\dfrac{3}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)
\(=\dfrac{-15}{24}+\dfrac{12}{24}\)
\(=\dfrac{-3}{24}\)
\(=-\dfrac{1}{8}\)
\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)
\(=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}\)
\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)
\(=-1+1-\dfrac{3}{7}\)
\(=-\dfrac{3}{7}\)
\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)
\(=\dfrac{6}{5}\times\dfrac{20}{7}\)
\(=\dfrac{120}{35}\)
\(=\dfrac{24}{7}\)
A=2^1+2^2+2^3+2^4+...+2^2010
=(2+2^2)+(2^3+2^4)+...+(2^2010+2^2011)
=2.(1+2)+2^3.(1+2)+...+2^2010.(1+2)
=2.3+2^3.3+...+2^2010.3
=(2+2^3+2^2010).3
=> A chia het cho 3
4:
a: =4/15-2,9+11/15=1-2,9=-1,9
b: \(=-36,75+3,7-63,25+6,3=10-100=-90\)
c: \(=6,5+3,5-\dfrac{10}{17}-\dfrac{7}{17}=10-1=9\)
d: \(=\dfrac{13}{25}\left(-39,1-60,9\right)=\dfrac{13}{25}\left(-100\right)=-52\)
e: =-5/12-7/12-3,7-6,3=-1-10=-11
f: =2,8(-6/13-7/13)-7,2=-2,8-7,2=-10
