a A B x y M N t F

Hai tia Ax và By cùng vuông góc với đường thẳng a do đó Ax // By.

Ta có: \(\widehat{AMN}+\widehat{MNB}=180^0\\ \Rightarrow\widehat{MNB}=180^0-\widehat{AMN}=180^0-120^0=60^0\)

B.

Gọi F là giao điểm của Mt và By ta có:

\(\widehat{AMF}+\widehat{MFB}=180^0\\ \Rightarrow\widehat{MFB}=180^0-\widehat{AMF}=180^0-90^0=90^0\\ \Rightarrow Mt\perp By.đpcm\)

`[x+4]/2018+[x+3]/2019=[x+2]/2020+[x+1]/2021`

`=>[x+4]/2018+1+[x+3]/2019+1=[x+2]/2020+1+[x+1]/2021+1`

`=>[x+2022]/2018+[x+2022]/2019-[x+2022]/2020-[x+2022]/2020=0`

`=>(x+2022)(1/2018+1/2019-1/2020-1/2022)=0`

    Mà `1/2018+1/2019-1/2020-1/2022`\(\ne 0\)

   `=>x+2022=0`

   `=>x=-2022`

Gọi số cần tìm là a

Ta có: 

\(\overline{a0}+a+\dfrac{\overline{a0}}{a}=175\\ \Leftrightarrow10a+a+\dfrac{10a+0}{a}=175\\ \Leftrightarrow11a+10=175\\ \Leftrightarrow11a=175-10=165\\ \Rightarrow a=165:11=15\)

Vậy số cần tìm là 15

Ta đặt: A=1.2+2.3+3.4+...+100.101

       =>3A=1.2.3+2.3.3+3.4.3+...+100.101.3

       =>3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101.(102-99)

       =>3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...+100.101.102

       =>3A=100.101.102

       =>3A=1030200

       =>A=343400

 Chúc bạn hok tôt :)!!!!!

\(\dfrac{x-5}{x+7}< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-5>0\\x+7< 0\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x-5< 0\\x+7>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>5\\x< -7\end{matrix}\right.\)  hoặc \(\left\{{}\begin{matrix}x< 5\\x>-7\end{matrix}\right.\)

\(\Leftrightarrow-7< x< 5\)

|x- 1/2| +2/3 = | 1/3 -2,5|

|x-1/2| + 2/3 = |-13/6|

| x -1/2|  + 2/3 = 13/6

| x-1/2| = 13/6 - 2/3

|x-1/2| = 3/2

x - 1/2 = + - 3/2

x = 2; x = - 1

vậy x ϵ {-1; 2}