Ta có : Tia OD nằm giữa hai tia OC và OB, nên ta có:

\(\widehat{COD}+\widehat{BOD}=\widehat{BOC}\)

Tia OC nằm giữa hai tia OD và OA, nên ta có:

\(\widehat{AOC}+\widehat{COD}=\widehat{AOD}\)

Mà \(\widehat{COD}=\widehat{COD}\) và \(\widehat{AOC}=\widehat{BOD}\) (theo giả thiết)

Nên \(\widehat{COD}+\widehat{BOD}=\widehat{AOC}+\widehat{COD}\)

Vậy \(\widehat{BOC}=\widehat{AOD}\) đpcm

Lê chí cường đúng 

\(-12\left(x-5\right)+7\left(3-x\right)=5\)

\(\Rightarrow-12x+60+21-7x=5\)

\(\Rightarrow-19x+81=5\)

\(\Rightarrow-19x=-76\Rightarrow x=4\)

Vậy \(x=4\)

Chúc học tốt !!!

3) \(A=\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}\)

\(=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\dfrac{2^{20}}{2^{12}}\)

\(=2^{20-12}=2^8\)

Vậy \(A=2^8\)

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

~ Học tốt ~

Đặt \(A=2^1+2^2+2^3+.....+2^{50}\)

\(\Rightarrow2A=2\left(2^1+2^2+2^3+....+2^{50}\right)\)

\(\Rightarrow2A=2^2+2^3+2^4+....+2^{51}\)

\(\Rightarrow2A-A=\left(2^2+2^3+2^4+....+2^{51}\right)-\left(2^1+2^2+2^3+....+2^{50}\right)\)

\(\Rightarrow A=2^{51}-2\)

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{100}}\)

\(\Rightarrow2A=2\times\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{100}}\right)\)

\(\Rightarrow2A=1+\dfrac{1}{2}+....+\dfrac{1}{2^{99}}\)

\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{100}}\right)\)\(\Rightarrow A=1-\dfrac{1}{2^{100}}\)

Vậy \(A=1-\dfrac{1}{2^{100}}\)

~ Học tốt ~

Ta có :

\(\dfrac{2014}{2015}+\dfrac{2015}{2016}+\dfrac{2016}{2014}=\left(1-\dfrac{1}{2015}\right)+\left(1-\dfrac{1}{2016}\right)+\left(1+\dfrac{2}{2014}\right)\) \(=\left(1+1+1\right)-\left(\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{2}{2014}\right)\)

\(=3-\left(\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{2}{2014}\right)\)

Dễ thấy : \(\left(\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{2}{2014}\right)>0\) vì \(\dfrac{1}{2015}>\dfrac{1}{2016}\)

Do đó \(\dfrac{2014}{2015}+\dfrac{2015}{2016}+\dfrac{2016}{2014}>3\)

~ Học tốt ~

2) a)

Ta có : \(16^{19}=\left(2^4\right)^{19}=2^{76}\)

\(8^{25}=\left(2^3\right)^{25}=2^{75}\)

Vì \(2^{76}>2^{75}\) nên \(16^{19}>8^{25}\)

Câu b cũng tương tự nha bạn làm cơ số bằng mũ ba là ra thui

~ Học tốt ~