CHO P=5^0+5^2+5^3+...+5^2016 và Q=5^2017:4
Tính Q-P
Cho P=5^0+5^1+5^2+...+5^2016 và Q=5^2017:4
Tính Q-P
Cho P=5^0 5^1 5^2 ... 5^2016 và Q=5^2017:4Tính Q-P
Cho P = 50 + 51 + 52 + ... + 52016 và Q = 52017 : 4
Tính: Q - P
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!
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
C = \(\frac{\left(a^{2016}+b^{2016}\right)^{2017}}{\left(c^{2016}+d^{2016}\right)^{2017}}\)= \(\frac{\left(a^{2017}-b^{2017}\right)^{2016}}{\left(c^{2017}-d^{2017}\right)^{2016}}\)
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
Rút gọn : B=(-5)\(^0\)+(-5)\(^1\)+(-5)\(^{^2}\)+(-5)\(^3\)+.......+(-5)\(^{2016}\)+(-5)\(^{2017}\)
-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
Rút gọn \(B=\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+...+\left(-5\right)^{2016}+\left(-5\right)^{2017}\)
\(B=\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2017}\)
\(-5B=\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+...+\left(-5\right)^{2017}\)
\(-6B=\left(-5\right)^{2017}-1\)
\(B=\frac{\left(-5\right)^{2017}-1}{-6}\)
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
\(-5B=\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2018}\)
\(-4B=\left(-5\right)^{2018}-\left(-5\right)^0\)
\(\Rightarrow B=\frac{\left(-5\right)^{2018}-\left(-5\right)^0}{-4}\)
M= 50+51+52+.....+52016 Q=52017:4 tính M-Q
\(2A=\frac{2^{2016}-4}{2^{2016}+1}=\frac{2^{2016}+1-5}{2^{2016}+1}=1-\frac{5}{2^{2016}+1}\)
\(2B=\frac{2^{2017}-4}{2^{2017}+1}=1-\frac{5}{2^{2017}+1}\)
Vì \(2^{2016} +1< 2^{2017}+1\)
=> \(2A< 2B\) Hay \(A< B\)
tìm số dư của 2015\(^{2016}\)\(^{2017}\) cho 5
đáp án : là 0 vì 5 chia hết cho 5