hi hivũ nguyễn hải châu bị nó lừa rồi  hay sao kia ha

2018² - 2017² + 2016² - 2015² + ... + 2² - 1²

= (2018+2017)(2018-2017) + (2016+2015)(2016-2015) + ... + (2+1)(2-1)

= 2018 + 2017 + 2016 + 2015 + ... + 2 + 1

= \(\dfrac{\left(1+2018\right).2018}{2}=2037171\)

\(\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)...\cdot\left(1-\dfrac{1}{1+2+3+...+2016}\right)\)

= \(\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)...\left(1-\dfrac{2}{2016.2017}\right)\)

= \(\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}...\dfrac{2014.2017}{2015.2016}.\dfrac{2015.2018}{2016.2017}\)

= \(\dfrac{2018}{3.2016}\)

= \(\dfrac{1009}{3024}\)

a , A = { 7 ; 8 ; 9 }

b , B = { 1 ;3; 5 7 ; 8; 10 ; 12 }

OA là tia phân giác của góc COE chứng minh:

Ta có : góc DOB =30o \(\Rightarrow\) góc EOA =30o(đối đỉnh)

Lại có :góc COA=30o

\(\Rightarrow\)góc EOA =góc COA (cùng bằng góc DOB)

\(\Rightarrow\)OA là tia phân giác của góc COE

a.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

=> \(\dfrac{4\left(bk\right)^4+5b^4}{4\left(dk\right)^4+5d^4}=\dfrac{b^4\left(4k^4+5\right)}{d^4\left(4k^4+5\right)}=\dfrac{b^4}{d^4}\)(1)

\(\dfrac{a^2b^2}{c^2d^2}=\dfrac{k^2b^2b^2}{k^2d^2d^2}=\dfrac{b^4}{d^4}\)(2)

Từ (1) và (2) suy ra: \(\dfrac{4a^4+5b^4}{4c^4+5d^4}=\dfrac{a^2b^2}{c^2d^2}\)

b.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

=> \(\dfrac{\left(bk\right)^{2004}-b^{2004}}{\left(bk\right)^{2004}+b^{2004}}=\dfrac{b^{2004}\left(k^{2004}-1\right)}{b^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\) (1)

\(\dfrac{\left(dk\right)^{2004}-d^{2004}}{\left(dk\right)^{2004}+d^{2004}}=\dfrac{d^{2004}\left(k^{2004}-1\right)}{d^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\) (2)

Từ (1) và (2) suy ra: \(\dfrac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\dfrac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\)

Xét: P² = \(\dfrac{\left(a-b\right)^2}{\left(a+b\right)^2}=\dfrac{a^2-2ab+b^2}{a^2+2ab+b^2}=\dfrac{3a^2+3b^2-6ab}{3a^2+3b^2+6ab}=\dfrac{10ab-6ab}{10ab+6ab}=\dfrac{4ab}{16ab}=\dfrac{1}{4}\)

=> P = \(\dfrac{1}{2}\)

Xét VP: (a³+b³)(a²+b²) - (a+b)

= a⁵ + b⁵ + a³b² + a²b³ - (a+b)

= a⁵ + b⁵ + a²b²(a+b) - (a+b)

= a⁵ + b⁵ + (a+b) - (a+b)

= a⁵ + b⁵ = VP (đpcm)

biết là sai nhưng ko biết làm chỉ biết kết quả

Ta có:

\(10^8+8=100000000+8=100000008⋮9\left(1\right)\)

mà \(10^8=2^8.5^8=2^3.2^5.5^8=8.2^5.5^8⋮8\)

\(\Rightarrow10^8+8⋮8\left(2\right)\)

Từ (1) và(2) => 10^8+8 chia hết cho 72