B=(-5)^0+(-5)^1+(-5)^2+(-5)^3+...+(-5)^2016+(-5)^2017
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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 = \(\dfrac{5^{2017}-1}{4}\)
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
\(B=1-5+5^2-5^3+...+5^{2016}-5^{2017}\) (1)
\(\Rightarrow5B=5-5^2+5^3-5^4+...+5^{2017}-5^{2018}\) (2)
Cộng vế với vế của (1) và (2):
\(6B=1+5-5+5^2-5^2+5^3-5^3+...+5^{2017}-5^{2017}-5^{2018}\)
\(\Rightarrow6B=1-5^{2018}\)
\(\Rightarrow B=\dfrac{1-5^{2018}}{6}\)
\(B=\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2017}\)
\(-5B=\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+...+\left(-5\right)^{2017}\)
\(-6B=\left(-5\right)^{2017}-1\)
\(B=\frac{\left(-5\right)^{2017}-1}{-6}\)
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
\(A=1+5+........+5^{2016}\)
\(\Leftrightarrow5A=5+5^2+.....+5^{2016}+2^{2017}\)
\(\Leftrightarrow5A-A=\left(5+5^2+......+5^{2017}\right)-\left(1+5+.....+5^{2016}\right)\)
\(\Leftrightarrow4A=5^{2017}-1\)
\(\Leftrightarrow A=\dfrac{5^{2017}-1}{4}\)
Mà \(B=\dfrac{5^{2017}}{4}\)
\(\Leftrightarrow B-A=\dfrac{5^{2017}}{4}-\dfrac{5^{2017}-1}{4}=1\)
Vậy....
\(P=5^0+5^1+5^2+...+5^{2016}\)
\(5P=5+5^2+...+5^{2017}\)
\(\Rightarrow5P-P=5^{2017}-1\)
\(\Rightarrow P=\frac{5^{2017}-1}{4}\)
\(\Rightarrow Q-P=\frac{5^{2017}}{4}-\frac{5^{2017}-1}{4}=\frac{1}{4}\)
P = 5^0+5^1+5^2+..+5^2016
5P = 5^1+5^2+5^3+...+5^2017
5P-P = (5^1+5^2+5^3+...+5^2017) - (5^0+5^1+5^2+..+5^2016)
4P = 5^1+5^2+5^3+...+5^2017-5^0-5^1-5^2-...-5^2016
4P = 5^2017-1
P = (5^2017-1):4
=> Q-P = 5^2017:4 - (5^2017-1) : 4
= (5^2017 - 5^2017 -1) : 4
= (-1) : 4 = -1/4
Vậy...
Chúc em học tốt!!!