\(d.4.4.4.4.4.16=2^2.2^2.2^2.2^2.2^2.2^4=2^{14}\)

\(e.5.5.5.5.3.3.3.15=5^4.3^3.3.5=5^5.3^4\)

\(e.6.6.6.7.7.7.42=6^3.7^3.2.3.6=2^3.3^3.7^4.2.3=2^4.3^4.7^4\)

4.4.4.4.4.16=4.4.4.4.4.4.4=47

5.5.5.5.3.3.3.15=15.15.15.5.15=154.5

6.6.6.7.7.7.42=42.42.42.42=424

\(d)4.4.4.4.4.16\\ =4.4.4.4.4.4.4\\ =4^7\)

\(e)5.5.5.5.3.3.3.15\\ =5.5.5.5.3.3.3.5.3\\ =5^5.3^4\)

\(e)6.6.6.7.7.7.42\\ =6.6.6.7.7.7.6.7\\ =6^4.7^4\)

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      =   \(\dfrac{5^{2017}-1}{4}\)

chấm hỏi lớn ???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

mega hỏi ?????

đè u đầu

chính xác lun

đè nát đầu

\(B=1-5+5^2-5^3+...+5^{2016}-5^{2017}\) (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

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

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

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

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

T = 1 -2 + 3 - 4 + 5 - .... +2015 - 2016 + 2017

T = (1 -2) + (3 - 4) + (5 - 6) +.......(2015 - 2016) + 2017

Mỗi nhóm có kết quả bằng  - 1 và số nhóm sẽ là: [(2016 - 1) + 1] : 2 = 1008

=> T = -1 x 1008 + 2017 = - 1008 + 2017 = 1009

T=1009

đúng 100% tick nha

nhin tren ghi vao cung doi k a con lau(nhung minh k rui)

Cách 1:
Xét số bị trừ, ta có:
(2/3 + 3/4 + 4/5 + ... + 2016/2017) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
= (2/3 + 3/4 + 4/5 + ... + 2015/2016 + 2016/2017) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
= (2/3 + 3/4 + 4/5 + ... + 2015/2016) x (1/2 + 2/3 + 3/4 + ... + 2015/2016) + 2016/2017 x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
Xét số trừ, ta có: 
(1/2 + 2/3 + 3/4 + ... + 2016/2017) x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= (1/2 + 2/3 + 3/4 + ... + 2015/2016 + 2016/2017) x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= (1/2 + 2/3 + 3/4 + ... + 2015/2016) x (2/3 + 3/4 + 4/5 + ... + 2015/2016) + 2016/2017 x (2/3 +3/4 + 4/5 + ... + 2015/2016) =
Ta thấy số bị trừ và số trừ có số hạng giống nhau là:
(2/3 +3/4 + 4/5 + ... + 2015/2016) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
Nên phép trừ trên có thể viết lại:
2016/2017 x (1/2 + 2/3 + 3/4 + ... + 2015/2016) - 2016/2017 x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= 2016/2017 x [(1/2 + 2/3 + 3/4 + ... + 2015/2016) - (2/3 +3/4 + 4/5 + ... + 2015/2016)]
= 2016/2017 x 1/2
= 1008/2017

Cách 2:

\(S=\dfrac{1}{2018!\left(2019-2018\right)!}+\dfrac{1}{2016!\left(2019-2016\right)!}+...+\dfrac{1}{2!\left(2019-2\right)!}+\dfrac{1}{0!\left(2019-0!\right)}\)

\(\Rightarrow2019!.S=\dfrac{2019!}{2018!\left(2019-2018\right)!}+\dfrac{2019!}{2016!\left(2019-2016\right)!}+...+\dfrac{2019!}{2!\left(2019-2\right)!}+\dfrac{2019!}{0!\left(2019-0\right)!}\)

\(=C_{2019}^{2018}+C_{2019}^{2016}+...+C_{2019}^2+C_{2019}^0\)

\(=\dfrac{1}{2}\left(C_{2019}^0+C_{2019}^1+...+C_{2019}^{2018}+C_{2019}^{2019}\right)\)

\(=\dfrac{1}{2}.2^{2019}=2^{2018}\)

\(\Rightarrow S=\dfrac{2^{2018}}{2019!}\)