a: \(1,\left(6\right)+\left(\dfrac{-2}{7}\right)-\left(-1,2\right)\)

\(=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\)

\(=\dfrac{175}{105}-\dfrac{30}{105}+\dfrac{126}{105}=\dfrac{271}{105}\)

b: \(0,\left(3\right)-\dfrac{-5}{6}+\dfrac{3}{4}=\dfrac{1}{3}+\dfrac{5}{6}+\dfrac{3}{4}\)

\(=\dfrac{4}{12}+\dfrac{10}{12}+\dfrac{9}{12}=\dfrac{23}{12}\)

c: \(0,\left(3\right)-1,\left(3\right)+\dfrac{2}{7}=\dfrac{1}{3}-\dfrac{4}{3}+\dfrac{2}{7}=-1+\dfrac{2}{7}=-\dfrac{5}{7}\)

d: \(-0,8\left(3\right)-\left(\dfrac{-3}{8}+\dfrac{1}{10}\right)\)

\(=-\dfrac{5}{6}+\dfrac{3}{8}-\dfrac{1}{10}\)

\(=-\dfrac{100}{120}+\dfrac{45}{120}-\dfrac{12}{120}=\dfrac{-67}{120}\)

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

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

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

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

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 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

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 \(\widehat{x'Ey}=130^0\)

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: \(\widehat{ACD}=\widehat{CDB}\)

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

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

5:

6:

Bài 5

a) Do Oc nằm giữa hai tia Oa và Ob nên

∠aOc + ∠cOb = ∠aOb

⇒ ∠cOb = ∠aOb - ∠aOc

= 100⁰ - 40⁰

= 60⁰

b) Do Od là tia phân giác của ∠cOb (gt)

⇒ ∠cOd = ∠cOb : 2

= 60⁰ : 2

= 30⁰

Xét 2 ΔABO và ΔADO ta có:

\(\widehat{BAO}=\widehat{DAO}\) (AD là phân giác của góc BAC) 

\(OA\) chung

\(\widehat{AOB}=\widehat{AOD}\left(gt\right)\) 

\(=>\Delta ABO=\Delta ADO\left(g.c.g\right)\) 

\(=>\widehat{B}=\widehat{D_1}\) (hai góc tương ứng)