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![](https://rs.olm.vn/images/avt/0.png?1311)
rối quá :)
B = (-5)0 + 51 + (-5)2 + 53 + ... + (-5)2016 + 52017
B = 1 + 51 + 52 + 53 + ... + 52016 + 52017
5B = 5 + 52 + 53 + ... + 52016 + 52017
5B - B = (5 + 52 + 53 + ... + 52016 + 52017) - (1 + 51 + 52 + 53 + ... + 52016 + 52017)
4B = 52017 - 1
B = \(\dfrac{5^{2017}-1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2017}\)
\(-5B=\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+...+\left(-5\right)^{2017}\)
\(-6B=\left(-5\right)^{2017}-1\)
\(B=\frac{\left(-5\right)^{2017}-1}{-6}\)
Ta có : B = (-5)^0 + (-5)^1 + ......+ (-5)^2017
(-5)B = (-5)^1 + (-5)^2 + .......+ (-5)^2018
(-4)B = (-5)^0- (-5)^2018
B = 1 - (-5)^2018 / (-4)
Nếu có sai sót gì mong các bạn bỏ qua
![](https://rs.olm.vn/images/avt/0.png?1311)
-5B=(-5)1+(-5)2+(-5)3+...+(-5)2018
-5B-B=[(-5)1+(-5)2+...+(-5)2018] - [(-5)0+(-5)1+...+(-5)2017]
-6B=(-5)2018-(-5)0 = (-5)2018-1
B= [(-5)2018-1]:-6
Anh học tốt nha ( em mới lớp 6)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=1-5+5^2-5^3+...+5^{2016}-5^{2017}\) (1)
\(\Rightarrow5B=5-5^2+5^3-5^4+...+5^{2017}-5^{2018}\) (2)
Cộng vế với vế của (1) và (2):
\(6B=1+5-5+5^2-5^2+5^3-5^3+...+5^{2017}-5^{2017}-5^{2018}\)
\(\Rightarrow6B=1-5^{2018}\)
\(\Rightarrow B=\dfrac{1-5^{2018}}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trục căn thức:
\(C=\frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+...+\)
\(+\frac{\left(\sqrt{2017}-\sqrt{2015}\right)}{\left(\sqrt{2017}+\sqrt{2015}\right)\left(\sqrt{2017}-\sqrt{2015}\right)}\)
\(C=\frac{\sqrt{3}-1}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+...+\frac{\sqrt{2017}-\sqrt{2015}}{2017-2015}\)
\(C=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+...+\frac{\sqrt{2017}-\sqrt{2015}}{2}\)
\(C=\frac{\sqrt{3}-1+\sqrt{5}-\sqrt{3}+...+\sqrt{2017}-\sqrt{2015}}{2}\)
\(C=\frac{\sqrt{2017}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Rút gọn:
\(A=5^0+5^1+5^2+...+5^{99}+5^{50}\)
\(5A=5^1+5^2+5^3+...+5^{51}\)
\(5A-A=\left(5^1+5^2+5^3+...+5^{51}\right)-\left(5^0+5^1+5^2+...+5^{50}\right)\)
\(4A=5^{51}-5^0\)
\(=>A=\left(5^{51}-5^0\right):4\)
Vậy : \(A=\left(5^{51}-5^0\right):4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+...+\left(-5\right)^{2017}\)
\(\Leftrightarrow-5B=\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+\left(-5\right)^4+...+\left(-5\right)^{2018}\)
\(\Leftrightarrow-5B-B=\left(-5\right)^{2018}-\left(-5\right)^0\)
\(\Leftrightarrow-6B=\left(-5\right)^{2018}-1\)
\(\Leftrightarrow B=\frac{\left(-5\right)^{2018}-1}{-6}\)
Bạn ơi vì sao ở dòng 3 lại là (-5)^2017 - (-5)^0 vậy??