Gọi F là giao điểm của Cy với AB

Bx//Cy

=>\(\widehat{BFC}=\widehat{xBC}\)(hai góc so le trong)

=>\(\widehat{BFC}=120^0\)

Ta có: \(\widehat{BFC}+\widehat{AFC}=180^0\)(hai góc kề bù)

=>\(\widehat{AFC}+120^0=180^0\)

=>\(\widehat{AFC}=60^0\)

Ta có: \(\widehat{ACF}+\widehat{ACy}=180^0\)(hai góc kề bù)

=>\(\widehat{ACF}+100^0=180^0\)

=>\(\widehat{ACF}=80^0\)

Xét ΔACF có \(\widehat{AFC}+\widehat{ACF}+\widehat{CAF}=180^0\)

=>\(\widehat{BAC}=180^0-60^0-80^0=40^0\)

kẻ CM//a và DN//bB(CM và Aa nằm cùng phía với nửa mặt phẳng chứa tia AC, DN và Bb nằm khác phía với nửa mặt phẳng chứa tia DB

CM//Aa

=>\(\widehat{MCA}=\widehat{A_1}\)

Ta có: CM//a

DN//b

mà a//b

nên CM//DN//a//b

CM//DN

=>\(\widehat{MCD}=\widehat{CDN}\)

DN//Bb

=>\(\widehat{NDB}=\widehat{B_1}\)

Ta có: \(\widehat{ACD}=\widehat{ACM}+\widehat{CDM}=\widehat{CDN}+\widehat{B_1}\)

\(\widehat{CDB}=\widehat{CDN}+\widehat{NDB}=\widehat{CDN}+\widehat{B_1}\)

Do đó: \(\widehat{ACD}=\widehat{CDB}\)

Bài 2:

Vì \(\widehat{xOz}< \widehat{xOy}\left(50^0< 80^0\right)\)

nên tia Oz nằm giữa hai tia Ox,Oy

=>\(\widehat{xOz}+\widehat{yOz}=\widehat{xOy}\)

=>\(\widehat{yOz}=80^0-50^0=30^0\)

Bài 4:

Ta có: \(\widehat{xEy}+\widehat{xEy'}=180^0\)(hai góc kề bù)

=>\(\widehat{xEy'}=180^0-50^0=130^0\)

Ta có: \(\widehat{xEy}=\widehat{x'Ey'}\)(hai góc đối đỉnh)

mà \(\widehat{xEy}=50^0\)

nên \(\widehat{x'Ey'}=50^0\)

Ta có: \(\widehat{xEy'}=\widehat{x'Ey}\)(hai góc đối đỉnh)

mà \(\widehat{xEy'}=130^0\)

nên \(\widehat{x'Ey}=130^0\)

\(\left|x-2\right|>=0\forall x\)

\(\left|2x+y-z\right|>=0\forall x,y,z\)

\(\left|2z+1\right|>=0\forall z\)

Do đó: \(\left|x-2\right|+\left|2x+y-z\right|+\left|2z+1\right|>=0\forall x,y,z\)

mà \(\left|x-2\right|+\left|2x+y-z\right|+\left|2z+1\right|< =0\)

nên Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\2x+y-z=0\\2z+1=0\\\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\z=-\dfrac{1}{2}\\y=-2x+z=-2\cdot2+\dfrac{-1}{2}=-4-\dfrac{1}{2}=-\dfrac{9}{2}\end{matrix}\right.\)

\(x^2+3x+1⋮x+1\)

=>\(x^2+x+2x+2-1⋮x+1\)

=>\(-1⋮x+1\)

=>\(x+1\in\left\{1;-1\right\}\)

=>\(x\in\left\{0;-2\right\}\)

bài 10: 

\(A=1+2012+2012^2+...+2012^{72}\)

=>\(2012A=2012+2012^2+...+2012^{73}\)

=>\(2012A-A=2012+2012^2+...+2012^{73}-1-2012-...-2012^{72}\)

=>\(2011A=2012^{73}-1\)

=>2011A=B

=>B>A

Bài 11:

\(B=\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\dfrac{3^{10}\left(11+5\right)}{3^9\cdot16}=\dfrac{3^{10}}{3^9}=3\)

\(C=\dfrac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}=\dfrac{2^{10}\left(13+65\right)}{2^8\cdot104}=\dfrac{2^2\cdot78}{104}=\dfrac{4\cdot2}{3}=\dfrac{8}{3}\)

mà \(3>\dfrac{8}{3}\)

nên B>C

6: \(\widehat{mOn}=\widehat{mOy}+\widehat{nOy}\)

\(=\dfrac{1}{2}\left(\widehat{xOy}+\widehat{zOy}\right)\)

\(=\dfrac{1}{2}\cdot180^0=90^0\)

5:

a: tia Oc nằm giữa hai tia Oa và Ob

=>\(\widehat{aOc}+\widehat{bOc}=\widehat{aOb}\)

=>\(\widehat{bOc}=100^0-40^0=60^0\)

b: Od là phân giác của góc cOb

=>\(\widehat{cOd}=\dfrac{\widehat{cOb}}{2}=\dfrac{60^0}{2}=30^0\)

Bài 7:

a: \(\left[0,\left(30\right)+0,\left(60\right)\right]x=10\)

=>\(\left(\dfrac{10}{33}+\dfrac{20}{33}\right)\cdot x=10\)

=>\(\dfrac{30}{33}\cdot x=10\)

=>\(x\cdot\dfrac{10}{11}=10\)

=>\(x=10:\dfrac{10}{11}=11\)

b: \(0,\left(12\right):1,\left(6\right)=x:0,\left(4\right)\)

=>\(x:\dfrac{4}{9}=\dfrac{4}{33}:\dfrac{5}{3}\)

=>\(x:\dfrac{4}{9}=\dfrac{4}{33}\cdot\dfrac{3}{5}=\dfrac{4}{11\cdot5}=\dfrac{4}{55}\)

=>\(x=\dfrac{4}{55}:\dfrac{4}{9}=\dfrac{9}{55}\)