7-2+2+3-7+2-1=
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\(A=1+7+7^2+7^3+...+7^{2007}\)
\(7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(7A-A=\left(7+7^2+7^3+7^4+...+7^{2008}\right)-\left(1+7+7^2+7^3+...+7^{2007}\right)\)
\(6A=7^{2008}-1\)
\(A=\frac{7^{2008}-1}{6}\)
Tương tự, \(B=\frac{4^{101}-1}{3},C=\frac{3^{101}-1}{2}\).
\(D=7+7^3+7^5+7^7+...+7^{99}\)
\(7^2.D=7^3+7^5+7^7+7^9+...+7^{101}\)
\(\left(7^2-1\right)D=\left(7^3+7^5+7^7+7^9+...+7^{101}\right)-\left(7+7^3+7^5+7^7+...+7^{99}\right)\)
\(48D=7^{101}-7\)
\(D=\frac{7^{101}-7}{48}\)
Tương tự, \(E=\frac{2^{9011}-2}{3}\)
a) -1/24 - [ 1/4 - ( 1/2 - 7/8 )]
= -1/24 - [ 1/4 +3/8 ]
= -1/24 - 5/8
= -2/3.
a) -1/24 - [ 1/4 - ( 1/2 - 7/8 )]
= -1/24 - [ 1/4 +3/8 ]
= -1/24 - 5/8
= -2/3.
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
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
5: \(=3-\dfrac{1}{4}+\dfrac{2}{3}-5+\dfrac{1}{3}+\dfrac{6}{5}-6+\dfrac{7}{4}-\dfrac{3}{2}\)
\(=3-5-6+\dfrac{-1}{4}+\dfrac{7}{4}+\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{6}{5}-\dfrac{3}{2}\)
\(=-8+\dfrac{3}{2}+1+\dfrac{-3}{10}\)
\(=-7+\dfrac{15-3}{10}=-7+\dfrac{6}{5}=-\dfrac{29}{5}\)
6: \(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=6-5-3-\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{3}+\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\)
\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)
7: \(=\dfrac{5}{3}-\dfrac{3}{7}+9-2-\dfrac{5}{7}+\dfrac{2}{3}+\dfrac{8}{7}-\dfrac{4}{3}-10\)
\(=9-2-10+\dfrac{5}{3}+\dfrac{2}{3}-\dfrac{4}{3}+\dfrac{-3}{7}-\dfrac{5}{7}+\dfrac{8}{7}\)
=-3+1
=-2
8: \(=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\)
\(=8+6-3+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}-1-\dfrac{2}{4}\)
\(=11+2-1-\dfrac{1}{2}\)
=11+1/2
=11,5
\(\left(\dfrac{5}{7}-\dfrac{7}{7}\right)-\left[0,2-\left(-\dfrac{2}{7}-\dfrac{1}{10}\right)\right]\)
=\(-\dfrac{2}{7}-\left[\dfrac{1}{5}+\dfrac{2}{7}+\dfrac{1}{10}\right]\)
=\(-\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{10}\)
=\(\left(-\dfrac{2}{7}-\dfrac{2}{7}\right)-\left(\dfrac{1}{5}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\left(\dfrac{2}{10}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\dfrac{3}{10}\)
=\(-\dfrac{40}{70}-\dfrac{21}{70}\)
=\(-\dfrac{61}{70}\)
(3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\)) - (5 - \(\dfrac{1}{3}\) - \(\dfrac{5}{6}\)) - (6 - \(\dfrac{7}{4}\) - \(\dfrac{3}{2}\))
= 3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\) - 5 + \(\dfrac{1}{3}\) + \(\dfrac{5}{6}\) - 6 + \(\dfrac{7}{4}\) + \(\dfrac{3}{2}\)
= (3 - 5 - 6) + ( \(\dfrac{7}{4}\) - \(\dfrac{1}{4}\)) + (\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)) + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= - 8 + \(\dfrac{3}{2}\) + 1 + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= (- 8 + 1) + (\(\dfrac{3}{2}\) + \(\dfrac{3}{2}\)) + \(\dfrac{5}{6}\)
= -7 + 3 + \(\dfrac{5}{6}\)
= - 4 + \(\dfrac{5}{6}\)
= \(\dfrac{-19}{6}\)
thôi chịu nhiều quá ai mà làm đc tự đi mà làm hỏi thì hỏi thì hỏi ít thôi người ta còn trả lời đc .
7 - 3 - 2 = 2 7 - 6 - 1 = 0 7 - 4 - 2 = 1
7 - 5 - 1 = 1 7 - 2 - 3 = 2 7 - 4 - 3 = 0
\(\dfrac{2}{7}:\dfrac{1}{4}-\dfrac{1}{7}=\dfrac{2}{7}x\dfrac{4}{1}-\dfrac{1}{7}=\dfrac{8}{7}-\dfrac{1}{7}=\dfrac{7}{7}=1\)
\(\dfrac{7}{11}x0+\dfrac{5}{9}:\dfrac{1}{2}=0+\dfrac{5}{9}x\dfrac{2}{1}=\dfrac{10}{9}\)
\(\left(\dfrac{3}{7}+\dfrac{1}{4}\right):\dfrac{3}{4}=\left(\dfrac{12}{28}+\dfrac{7}{28}\right)x\dfrac{4}{3}=\dfrac{19}{28}x\dfrac{4}{3}=\dfrac{19}{21}\)
\(\dfrac{4}{3}x\dfrac{1}{2}+\dfrac{7}{2}:\dfrac{1}{4}=\dfrac{4}{6}+\dfrac{7}{2}x\dfrac{4}{1}=\dfrac{2}{3}+\dfrac{14}{1}=\dfrac{2}{3}+14=14\dfrac{2}{3}=\dfrac{44}{3}\)
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