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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, \(E=\frac{2^{9011}-2}{3}\)
2) -3(4 - 7) + 5(-3 + 2)
= -3.4 + 3.7 - 5.3 + 5.2
= -12 + 21 -15 + 10
= 31 - 27
= 4
4) -5(2 - 7) + 4(2 - 5)
= -5.2 + 5.7 + 4.2 - 4.5
= -10 + 35 + 8 - 20
= 38 - 30
= 8
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
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
3: \(=20-12-8+12=20-8=12\)
5: \(=-18-42-21-35=-116\)
3: \(=-15+18-12+8=-27+26=-1\)
2: \(=-12+21-15+10=9-5=4\)
Ta có
A = \(\dfrac{1+7+7^2+7^3+...+7^{11}}{1+7+7^2+7^3+...+7^{10}}\)
Đặt C = 1 + 7 + 72 + 73+...+711
7C = 7 + 72 + 73 + ... + 711 + 712
=> 6C = 712 - 1
C = \(\dfrac{7^{12}-1}{6}\)
Đặt D = 1 + 7 + 72 + 73+...+710
7D = 7 + 72 + 73 + ... + 710 + 711
=> 6D = \(7^{11}-1\)
D = \(\dfrac{7^{11}-1}{6}\)
=> A = \(\dfrac{\dfrac{7^{12}-1}{6}}{\dfrac{7^{11}-1}{6}}\)
A = \(\dfrac{7^{12}-1}{6}\) : \(\dfrac{7^{11}-1}{6}\)
A = \(\dfrac{7^{12}-1}{6}.\dfrac{6}{7^{11}-1}\)
A = \(\dfrac{7^{12}-1}{7^{11}-1}\) = 7, 000000003
Lại có:
B = \(\dfrac{1+3+3^2+3^3+...+3^{11}}{1+3+3^2+3^3+...+3^{10}}\)\
Đặt H = \(1+3+3^2+3^3+...+3^{11}\)
3H = \(3+3^2+3^3+...+3^{12}\)
=> 2H = \(3^{12}-1\)
H = \(\dfrac{3^{12}-1}{2}\)
Đặt Q = \(1+3+3^2+3^3+...+3^{10}\)
3Q = \(3+3^2+3^3+...+3^{10}+3^{11}\)
=> 2Q = \(3^{11}-1\)
Q = \(\dfrac{3^{11}-1}{2}\)
=> B = \(\dfrac{\dfrac{3^{12}-1}{2}}{\dfrac{3^{11}-1}{2}}\)
B = \(\dfrac{3^{12}-1}{2}:\dfrac{3^{11}-1}{2}\)
B = \(\dfrac{3^{12}-1}{2}.\dfrac{2}{3^{11}-1}\)
B = \(\dfrac{3^{12}-1}{3^{11}-1}\)
B = 3, 00001129
Vì 7, 000000003 > 3, 00001129
=> A > B
Vậy A > B
\(\left(\dfrac{5}{7}-\dfrac{7}{7}\right)-\left[0,2-\left(-\dfrac{2}{7}-\dfrac{1}{10}\right)\right]\)
=\(-\dfrac{2}{7}-\left[\dfrac{1}{5}+\dfrac{2}{7}+\dfrac{1}{10}\right]\)
=\(-\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{10}\)
=\(\left(-\dfrac{2}{7}-\dfrac{2}{7}\right)-\left(\dfrac{1}{5}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\left(\dfrac{2}{10}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\dfrac{3}{10}\)
=\(-\dfrac{40}{70}-\dfrac{21}{70}\)
=\(-\dfrac{61}{70}\)
(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 .
a) 8697-(37:35+2(13-3)
=8697-(32+2.13-2.3)
=8697-(9+26-6)
=8697-29
=8668
b) 2010-2000:(486-2(72-6))
=2010-2000:(486-(2.49-2.6))
=2010-2000:(486-(98-12)
=2010-2000:400
=2010-5
=2005
c)718:716+22.33.32.5-22.7+83
=(718:716)+(22.33.32)-(22.7)+83
=49+972-28+83
=1076
nhớ tích nha bạn ^-^