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\(S=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+....+\left(-\dfrac{1}{7}\right)^{2017}\\ =1+-\dfrac{1}{7}+\dfrac{1}{7^2}+-\dfrac{1}{7^3}+.....+-\dfrac{1}{7^{2017}}\\ =\left(1+\dfrac{1}{7^2}+\dfrac{1}{7^4}+...+\dfrac{1}{7^{2016}}\right)-\left(\dfrac{1}{7}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{2017}}\right)\)
rồi bạn tính 2 về rồi trừ ra là xng nhé
\(S=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2017}\\ S=\dfrac{\left(-1\right)^0}{7^0}+\dfrac{\left(-1\right)^1}{7^1}+\dfrac{\left(-1\right)^2}{7^2}+...+\dfrac{\left(-1\right)^{2017}}{7^{2017}}\\ S=\dfrac{1}{7^0}+\dfrac{-1}{7^1}+\dfrac{1}{7^2}+...+\dfrac{-1}{7^{2017}}\\ -7S=\dfrac{-7}{7^0}+\dfrac{7}{7^1}+\dfrac{-7}{7^2}+...+\dfrac{7}{7^{2017}}\\ -7S=\left(-7\right)+\dfrac{1}{7^0}+\dfrac{-1}{7^1}+...+\dfrac{1}{7^{2016}}\\ -7S-S=\left[\left(-7\right)+\dfrac{1}{7^0}+\dfrac{-1}{7^1}+...+\dfrac{1}{7^{2016}}\right]+\left(\dfrac{1}{7^0}+\dfrac{-1}{7^1}+\dfrac{1}{7^2}+...+\dfrac{-1}{7^{2017}}\right)\\ -8S=\left(-7\right)+\dfrac{-1}{2017}\\ -8S=-\left(7+\dfrac{1}{2017}\right)\\ 8S=7+\dfrac{1}{2017}\\ S=\dfrac{7+\dfrac{1}{2017}}{8}\)
Vậy ...
S= \(\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2017}\)
\(\left(-\dfrac{1}{7}\right)S=\left(-\dfrac{1}{7}\right)\left(-\dfrac{1}{7}+-\dfrac{1^2}{7}+..+-\dfrac{1^{2007}}{7}\right)\)
= \(-\dfrac{1}{7}+-\dfrac{1}{7}^2+....+\dfrac{-1^{2008}}{7}\)
=>\(-\dfrac{1}{7}S-S=\) \(-\dfrac{1}{7}+-\dfrac{1}{7}^2+....+\dfrac{-1^{2008}}{7}\) \(-\)\(\left(1+-\dfrac{1}{7}+-\dfrac{1^2}{7}+...+-\dfrac{1^{2007}}{7}\right)\)
=> \(-\dfrac{1}{7}S=\) \(\dfrac{-1^{2008}}{7}-1\)
=> S= \(\dfrac{-1^{2008}}{7}-1\) : \(\dfrac{-1}{7}\)
a) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5=1+1+0,5=2,5\)b)
\(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{7}{7}.33\dfrac{1}{3}=\dfrac{7}{3}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{7}{3}.\left(-14\right)=-\dfrac{1}{6}\)
c,
\(\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{5}{7}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-5}{7}\right)=\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{7}{5}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-7}{5}\right)\)
\(\left(-\dfrac{7}{5}\right)\left(15\dfrac{1}{4}+2010-25\dfrac{1}{4}-2016\right)=\left(-\dfrac{7}{5}\right)\left(-10-6\right)=22,4\)
d,
\(\left(2017-\dfrac{3}{7}+\dfrac{9}{11}\right)-\left(2016-\dfrac{3}{7}+\dfrac{8}{17}\right)-\left(2015+\dfrac{9}{11}-\dfrac{8}{17}\right)=2017-\dfrac{3}{7}+\dfrac{9}{11}-2016+\dfrac{3}{7}-\dfrac{8}{17}-2015-\dfrac{9}{11}+\dfrac{8}{17}\)\(\left(2017-2016-2015\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(-\dfrac{8}{17}+\dfrac{8}{17}\right)=-2014\)
Bạn ơi cho mình hỏi tại sao đề bài câu c là -5/7 mà bn lm -7/5
Các bạn trả lời giúp mk nha. Mk đang cần gấp. Chều nay mk kiểm tra rồi
1: \(\left(\dfrac{1}{2}\right)^{27}=\left(\dfrac{1}{8}\right)^9\)
\(\left(\dfrac{1}{3}\right)^{18}=\left(\dfrac{1}{9}\right)^9\)
mà 1/8>1/9
nên \(\left(\dfrac{1}{2}\right)^{27}>\left(\dfrac{1}{3}\right)^{18}\)
2: \(\dfrac{-4}{7}=\dfrac{-32}{56}\)
\(\dfrac{-5}{8}=\dfrac{-35}{56}\)
mà -32>-35
nên \(\dfrac{-4}{7}>\dfrac{-5}{8}\)
hay \(\left(\dfrac{-4}{7}\right)^{205}>\left(-\dfrac{5}{8}\right)^{205}\)
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
B = .................
Xét thừa số 63.1,2 - 21.3,6 = 0 nên B = 0
\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{2\left(\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}\right)}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{1}{2}\right|=\dfrac{1}{18}+\dfrac{9}{10}=\dfrac{43}{45}\)
Mình làm câu 1,2 trước, câu 3 sau
Câu 1:
\(\sqrt{x^2}=0\)
=> \(\left(\sqrt{x^2}\right)^2=0^2\)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Câu 2:
\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)
\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)
mk ko chép đề đâu nha
\(S=1+\dfrac{-1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{2016}}\)
đặt \(7S=7-1+\dfrac{1}{7}+...+\dfrac{1}{7^{2015}}\)
=>\(7S+S=\left(7-1+\dfrac{1}{7}+...+\dfrac{1}{7^{2015}}\right)+\left(1-\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{2016}}\right)\)
=>\(8S=7-1+\dfrac{1}{7}+...+\dfrac{1}{7^{2015}}+1-\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{2016}}\)
=>\(8S=7+\left(-1+1\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+...+\left(\dfrac{1}{7^{2015}}-\dfrac{1}{7^{2015}}\right)+\dfrac{1}{7^{2016}}\)
=> \(8S=7+\dfrac{1}{7^{2016}}\)
\(\Rightarrow S=\dfrac{7+\dfrac{1}{7^{2016}}}{8}\)
Gỉa sử : \(-\dfrac{1}{7}=a\)
Thay vào S ,có :
\(a^0+a^1+a^{2^{ }}+.........+a^{2016}\) (1)
=> a.S = a( \(a^0+a^1+a^{2^{ }}+.........+a^{2016}\) )
= \(a^1+a^2+a^3+.........+a^{2016}+a^{2017}\) (2)
Lấy (2) - (1) ,CÓ :
aS-S = a2017 -1 => S(a-1) = a2017 -1
=> S = \(\dfrac{a^{2017}-1}{a-1}\)
Thay a= -1/7 vào S = \(\dfrac{a^{2017}-1}{a-1}\) ,có :
S = \(\dfrac{\left(\dfrac{-1}{7}\right)^{2017}-1}{-\dfrac{1}{7}-1}=\dfrac{\left(-\dfrac{1}{7}\right)^{2017}}{-\dfrac{8}{7}}\)