Vì \(\widehat{AOB}\)và \(\widehat{BOC}\)kề bù nên \(\widehat{AOB}\)+\(\widehat{BOC}\)=\(180^0\)\(\Rightarrow5\widehat{AOB}\)+\(\widehat{AOB}\)=\(180^0\)\(\Rightarrow6\widehat{AOB}\)=\(180^0\)                                                                                                                                                                   \(\Rightarrow\widehat{AOB}\)\(180^0:6=30^0\)

Vì \(\widehat{AOB}\)+\(\widehat{BOC}\)=\(180^0\)\(\Rightarrow30^0+\widehat{BOC}=180^0\)\(\Rightarrow\widehat{BOC}\)\(=180^0-30^0=150^0\)

b,Vì OD là phân giác của \(\widehat{BOC}\)\(\Rightarrow\)OD nằm giữa và \(\widehat{COD}=\widehat{DOB}=\frac{\widehat{BOC}}{2}=\frac{150^0}{2}=75^0\)

Vì \(\widehat{DOB}=75^0>30^0=\widehat{AOB}\)

Trên cùng nửa mặt phẳng bờ OA có \(\widehat{AOB}< \widehat{DOB}\Rightarrow OB\)nằm giữa \(OA\)và \(OD\)

                                                                                      \(\Rightarrow\widehat{DOB}+\widehat{AOB}=\widehat{AOD}\)

                                                                                        \(\Rightarrow75^0+30^0=\widehat{AOD}\)

                                                                                         \(\Rightarrow\widehat{AOD}=100^0\)

Phần c tự làm nhé

Học tok

Theo đề bài,ta có :

  A = \((1+3^2)+(3^4+3^6+3^8)+...+(3^{2002}+3^{2004}+3^{2006})\)

  A = \(10+3^4(1+3^2+3^4)+...+3^{2002}(1+3^2+3^4)\)

  A = \(10+3^4\cdot91+...+3^{2002}\cdot91\)

  A = \(10+(3^4+...+3^{2002})\cdot91\)

  A = \(10+7\cdot13(3^4+...+3^{2002})\)

Vậy : \(A=1+3^2+3^4+3^6+...+3^{2004}+3^{2006}⋮13\)dư 10

Chúc bạn học tốt

x = -1 và x = -5

\(\frac{7}{5}-\frac{1}{5}.\left|x+3\right|=\)

\(\frac{1}{5}.\left|x+3\right|=\frac{7}{5}-1\)

\(\frac{1}{5}.\left|x+3\right|=\frac{2}{5}\)

\(\left|x+3\right|=\frac{2}{5}:\frac{1}{5}\)

\(\left|x+3\right|=\frac{2}{5}.5\)

\(\left|x+3\right|=2\)

B tự làm nốt nha

Đề sai đúng không đáng lẽ phải như này

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\left(đpcm\right)\)

\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{40}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\left(đpcm\right)\)

\(2n^2+9⋮n-2\)

\(\Leftrightarrow2n\left(n-2\right)+4n+9⋮n-2\)

-Mà: \(2n\left(n-2\right)⋮n-2\Rightarrow4n+9⋮n-2\)

\(\Leftrightarrow4\left(n-2\right)+17⋮n-2\)

\(\Rightarrow17⋮n-2\Leftrightarrow n-2\inƯ\left(17\right)\)

....

\(\frac{5}{4}+\left(\frac{7}{3}\times y-1\right)\div\frac{3}{2}=6\)

\(\frac{5}{4}+\left(\frac{7}{3}\times y-1\right)=6\times\frac{3}{2}\)

\(\frac{5}{4}+\left(\frac{7}{3}\times y-1\right)=9\)

\(\frac{7}{3}\times y-1=9-\frac{5}{4}\)

\(\frac{7}{3}\times y-1=\frac{31}{4}\)

\(\frac{7}{3}\times y=\frac{31}{4}+1\)

\(\frac{7}{3}\times y=\frac{35}{4}\)

\(y=\frac{35}{4}\div\frac{7}{3}\)

\(y=\frac{35}{4}\times\frac{3}{7}\)

\(y=\frac{5}{4}\times\frac{3}{1}\)

\(y=\frac{15}{4}\)

\(2n+3\)và \(3n+4\)

Gọi d là ước chung lớn nhất của \(2n+3\)và \(3n+4\)

Ta có :

\(2n+3⋮d=\left(2n+3\right)\cdot3⋮d=\left(6n+9\right)⋮d\)

\(3n+4⋮d=\left(3n+4\right)\cdot2⋮d=\left(6n+8\right)⋮d\)

\(\Rightarrow\left(6n+9\right)-\left(6n+8\right)⋮d\)

\(\Rightarrow6n+9-6n-8⋮d\)

\(\Rightarrow1⋮d\)

\(\Rightarrow d=1\)

\(\)Vậy \(2n+3\)và \(3n+4\)là hai số nguyên tố cùng nhau

Gọi ƯCLN ( 2n+3;3n+4 ) là d

\(\Rightarrow\orbr{\begin{cases}2n+3⋮d\\3n+4⋮d\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3.\left(2n+3\right)⋮d\\2.\left(3n+4\right)⋮d\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}6n+9⋮d\\6n+8⋮d\end{cases}}\)\(\Rightarrow\left(6n+9\right)-\left(6n+8\right)⋮d\)

\(\Rightarrow1⋮d\)

\(\Rightarrow d\in\text{Ư}\left(1\right)=\pm1\)

\(\Rightarrow\)2n+3 và 3n+4 là 2 số nguyên tố cùng nhau

                                                đpcm