\(P=5^0+5^1+5^2+...+5^{2016}\)

\(5P=5+5^2+...+5^{2017}\)

\(\Rightarrow5P-P=5^{2017}-1\)

\(\Rightarrow P=\frac{5^{2017}-1}{4}\)

\(\Rightarrow Q-P=\frac{5^{2017}}{4}-\frac{5^{2017}-1}{4}=\frac{1}{4}\)

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!!!

Q - P = 5²⁰¹⁷ : 4 - (5²⁰¹⁷-1):4

         = (5²⁰¹⁷ - 5²⁰¹⁷ + 1) : 4 

         = 1 : 4 = 1/4

Vậy...

Nãy mình quên đổi dấu!!!

P = 5⁰ + 5¹ + 5² + .... + 5²⁰¹⁶

5P = 5( 5⁰ + 5¹ + 5² + .... + 5²⁰¹⁶ )

= 5¹ + 5² + 5³ + .... + 5²⁰¹⁷ 

5P - P = ( 5¹ + 5² + 5³ + .... + 5²⁰¹⁷ ) - ( 5⁰ + 5¹ + 5² + .... + 5²⁰¹⁶ )

4P = 5²⁰¹⁷ - 1

=> P = ( 5²⁰¹⁷ - 1 ) : 4

=> Q - P = 5²⁰¹⁷ : 4 - ( 5²⁰¹⁷ - 1 ) : 4

= [ 5²⁰¹⁷ - ( 5²⁰¹⁷ - 1) ] : 4

= ( 5²⁰¹⁷ - 5²⁰⁴⁷ + 1 ) : 4

= 1/4

\(\frac{1}{4}\)nhớ ấn đúng cho mình nhé!

P < Q

Tk mk nha, thề 100% đúng

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 = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)

\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)

\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)

 \(=\frac{1}{3.2}-\frac{5.2}{7.3}\)

\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)

\(=\frac{7}{42}-\frac{20}{42}\)

\(=-\frac{13}{42}\)

cs ng làm đung r

đag định lm

-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)

Cho e sửa lại dòng cuối :

B= [(-5)²⁰¹⁸-(-1)]:-6

\(A=\frac{5^{2016}+1}{5^{2017}+1}\)

\(\Rightarrow5A=\frac{5^{2017}+5}{5^{2017}+1}=1+\frac{4}{5^{2017}+1}\)

\(B=\frac{5^{2017}+1}{5^{2018}+1}\)

\(\Rightarrow5B=\frac{5^{2018}+5}{5^{2018}+1}=1+\frac{4}{5^{2018}+1}\)

Do \(\frac{4}{5^{2018}+1}< \frac{4}{5^{2017}+1}\)

\(\Rightarrow5A>5B\Leftrightarrow A>B\)

\(B=\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2017}\)

\(-5B=\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+...+\left(-5\right)^{2017}\)

\(-6B=\left(-5\right)^{2017}-1\)

\(B=\frac{\left(-5\right)^{2017}-1}{-6}\)

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

\(-5B=\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2018}\)

\(-4B=\left(-5\right)^{2018}-\left(-5\right)^0\)

\(\Rightarrow B=\frac{\left(-5\right)^{2018}-\left(-5\right)^0}{-4}\)