đố mẹo

11 + 16 = 27

cái câu hỏi này vẫn tính như bình thường thôi

làm sao mà có thể làm khó bộ não thông minh của tôi được

:)))))))))))))

Lấy cái gì để chọn em

\(\dfrac{5^2}{1\cdot6}+\dfrac{5^2}{6\cdot11}+...+\dfrac{5^2}{26\cdot31}\)

\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)

\(=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)

\(=5\left(1-\dfrac{1}{31}\right)=5\cdot\dfrac{30}{31}=\dfrac{150}{31}\)

Tớ không biết

\(25M=\dfrac{5^{12}+25}{5^{12}+1}=1+\dfrac{24}{5^{12}+1}\)

\(25N=\dfrac{5^{20}}{5^{20}+1}=\dfrac{5^{20}+1-1}{5^{20}+1}=1-\dfrac{1}{5^{20}+1}\)

\(\dfrac{24}{5^{12}+1}>\dfrac{-1}{5^{20}+1}\)

=>\(\dfrac{24}{5^{12}+1}+1>\dfrac{-1}{5^{20}+1}+1\)

=>25M>25N

=>M>N

a) Đặt A = 10.11 + 11.12 + 12.13 + ... + 86.86

⇒ 3A = 10.11.3 + 11.12.3 + 12.13.3 + ... + 85.86.3

= 10.11.(12 - 9) + 11.12.(13 - 10) + 12.13.(14 - 11) + ... + 85.86.(87 - 84)

= 10.11.12 - 9.10.11 + 11.12.13 - 10.11.12 + 12.13.14 - 11.12.13 + ... + 85.86.87 - 84.85.86

= -9.10.11 + 85.86.87

= 634980

⇒ A = 634980 : 3 = 211660

b) Đặt B = 1.4 + 4.7 + 7.10 + ... + 37.40 + 40.43

⇒ 9B = 1.4.9 + 4.7.9 + 7.10.9 + ... + 37.40.9 + 40.43.9

= 1.4.(7 + 2) + 4.7.(10 - 1) + 7.10.(13 - 4) + ... + 37.40.(43 - 34) + 40.43.(46 - 37)

= 1.4.7 + 1.2.4 + 4.7.10 - 1.4.7 + 7.10.13 - 4.7.10 + ... + 37.40.43 - 34.37.40 + 40.43.46 - 37.40.43

= 1.2.4 + 40.43.46

= 8 + 79120

= 79128

⇒ B = 79128 : 9 = 8792

Gọi d=ƯCLN(2n+1;3n+2)

=>\(\left\{{}\begin{matrix}2n+1⋮d\\3n+2⋮d\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}6n+3⋮d\\6n+4⋮d\end{matrix}\right.\)

=>\(6n+4-6n-3⋮d\)

=>\(1⋮d\)

=>d=1

=>ƯCLN(2n+1;3n+2)=1

=>\(\dfrac{2n+1}{3n+2}\) là phân số tối giản

Đặt \(B=4^{2023}+4^{2022}+...+4^2+5\)

=>\(B=4^{2023}+4^{2022}+...+4^2+4+1\) và \(A=75B+25\)

\(B=4^{2023}+4^{2022}+...+4^2+4+1\)

=>\(4B=4^{2024}+4^{2023}+...+4^3+4^2+4\)

=>\(4B-B=4^{2024}+4^{2023}+...+4^3+4^2+4-4^{2023}-4^{2022}-...-4^2-4-1\)

=>\(3B=4^{2024}-1\)

=>\(B=\dfrac{4^{2024}-1}{3}\)

\(A=75\cdot B+25=75\cdot\dfrac{4^{2024}-1}{3}+25\)

\(=25\left(4^{2024}-1\right)+25\)

\(=25\cdot4^{2024}⋮4^{2024}\)