![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left\{{}\begin{matrix}Az\perp Ox\\Ox\perp Oy\left(\widehat{xOy}=90^0\right)\end{matrix}\right.\Rightarrow Az//Oy\)
\(b,\widehat{xOm}=\dfrac{1}{2}\widehat{xOy}=\dfrac{1}{2}\cdot90^0=45^0\left(t/c.phân.giác\right)\\ \widehat{nAx}=\dfrac{1}{2}\widehat{xAz}=\dfrac{1}{2}\cdot90^0=45^0\left(t/c.phân.giác\right)\\ \Rightarrow\widehat{xOm}=\widehat{nAx}\left(=45^0\right)\)
Mà 2 góc này ở vị trí đồng vị nên \(Om//An\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì BC và Cx là 2 tia đối nên \(\widehat{BCA}\) và \(\widehat{ACx}\) là 2 góc kề bù
\(\Rightarrow\widehat{ACB}+\widehat{ACx}=180^o\)
\(40^o+\widehat{ACx}=180^o\)
\(\widehat{ACx}=140^o\)
b) Ta có:\(\widehat{ACB}+\widehat{ABC}+\widehat{BAC}=180^o\) (tổng 3 góc trong 1 tam giác)
\(40^o+\widehat{ABC}+70^o=180^o\)
\(\widehat{ABC}=70^o\)(1)
Vì Oy là phân giác của \(\widehat{ACx}\) nên \(\widehat{xCy}=\dfrac{\widehat{ACx}}{2}=\dfrac{140^o}{2}=70^o\)(2)
Từ (1),(2) => \(\widehat{ABC}=\widehat{xCy}\)
c)Cặp góc đồng vị là \(\widehat{ABC}\) và \(\widehat{xCy}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt \(A=2+2^2+2^3+...+2^{2018}\)
\(\Rightarrow2A=2^2+2^3+2^4+...+2^{2019}\)
\(\Rightarrow2A-A=2^{2019}-2\)
\(\Rightarrow A=2^{2019}-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
2:
a: =>x-5=0 hoặc 2x-4=0
=>x=2 hoặc x=5
b: =>4x-4=2x-6x+12
=>4x-4=-4x+12
=>8x=16
=>x=2
c: =>9-x+x-6=3-x+12
=>15-x=3
=>x=12
d: =>-152-(3x+1)=54
=>3x+1=-152-54=-206
=>3x=-207
=>x=-69
Bài tập 1:
a) \(12x+2^4=4x-24\)
\(\Rightarrow12x+16=4x-24\)
\(\Rightarrow12x-4x=-24-16\)
\(\Rightarrow8x=-40\)
\(\Rightarrow x=\dfrac{-40}{8}=-5\)
b) \(6\left(4-x\right)-4\left(x-1\right)=2x+40\)
\(\Rightarrow24-6x-4x+4=2x+40\)
\(\Rightarrow28-10x=2x+40\)
\(\Rightarrow-10x-2x=40-28\)
\(\Rightarrow-12x=12\)
\(\Rightarrow x=-\dfrac{12}{12}=-1\)
c) \(12\left(2x+7\right)-9\left(3x-2\right)=72\)
\(\Rightarrow24x+84-27x+18=72\)
\(\Rightarrow24x-27x=72-84-18\)
\(\Rightarrow-3x=-30\)
\(\Rightarrow x=-\dfrac{30}{-3}=10\)
d) \(6\left(x-2\right)=20-4x\)
\(\Rightarrow6x-12=20-4x\)
\(\Rightarrow6x+4x=20+12\)
\(\Rightarrow10x=32\)
\(\Rightarrow x=\dfrac{32}{10}=\dfrac{16}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: Ta có: \(47\dfrac{1}{9}:\left(-\dfrac{5}{2}\right)-27\dfrac{1}{9}:\left(-\dfrac{5}{2}\right)\)
\(=\left(47+\dfrac{1}{9}\right)\cdot\dfrac{-2}{5}-\left(27+\dfrac{1}{9}\right)\cdot\dfrac{-2}{5}\)
\(=20\cdot\dfrac{-2}{5}\)
=-8