Bài 11: \(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\)

=>\(\left(2a+13b\right)\left(3c-7d\right)=\left(3a-7b\right)\left(2c+13d\right)\)

=>\(6ac-14ad+39bc-91bd=6ac+39ad-14bc-91bd\)

=>-14ad-39ad=-14bc-39bc

=>ad=bc

=>\(\dfrac{a}{b}=\dfrac{c}{d}\)

Bài 12:

\(\dfrac{a+2019}{a-2019}=\dfrac{b+2020}{b-2020}\)

=>\(\left(a+2019\right)\left(b-2020\right)=\left(a-2019\right)\left(b+2020\right)\)

=>\(ab-2020a+2019b-2019\cdot2020=ab+2020a-2019b-2019\cdot2020\)

=>-2020a-2020a=-2019b-2019b

=>2020a=2019b

=>\(\dfrac{a}{2019}=\dfrac{b}{2020}\)

\(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(x-4\right)-\left(x^2-16\right)\left(x+4\right)+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(x-4-x-4+3\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

Gọi tuổi mẹ là x(tuổi)

(Điều kiện: x>0)

Tuổi con là \(\dfrac{3}{5}x\left(tuổi\right)\)

Tuổi mẹ cách đây 9 năm là x-9(tuổi)

Tuổi con cách đây 9 năm là \(\dfrac{3}{5}x-9\left(tuổi\right)\)

Tuổi mẹ gấp đôi tuổi con nên \(x-9=2\left(\dfrac{3}{5}x-9\right)\)

=>\(\dfrac{6}{5}x-18=x-9\)

=>\(\dfrac{1}{5}x=9\)

=>x=45(nhận)

vậy: Tuổi mẹ là 45 tuổi

Tuổi con là \(\dfrac{3}{5}\cdot45=27\left(tuổi\right)\)

Bài 3:

\(a.-4,36\\ =\dfrac{-436}{100}\\ =\dfrac{-436:4}{100:4}\\ =\dfrac{-109}{25}\\b.0,708\\ =\dfrac{708}{1000}\\ =\dfrac{708:4}{1000:4}\\ =\dfrac{177}{250}\)

Bài 4: \(\dfrac{5}{12}=0,41\left(6\right)\)

\(-\dfrac{8}{11}=-0,\left(72\right)\)

\(\dfrac{3}{22}=0,1\left(36\right)\)

\(-\dfrac{111}{36}=-3,08\left(3\right)\)

Bài 5:

a: \(3,\left(15\right)=3+\dfrac{15}{99}=3+\dfrac{5}{33}=\dfrac{3\cdot33+5}{33}=\dfrac{104}{33}\)

b: \(0,2\left(07\right)=0,2+0,0\left(07\right)=\dfrac{41}{198}\)

c: \(0,1\left(37\right)=0,1+0,0\left(37\right)=\dfrac{1}{10}+\dfrac{37}{990}=\dfrac{68}{495}\)

d: \(0,20\left(23\right)=0,20+0,00\left(23\right)=0,2+\dfrac{23}{9900}=\dfrac{2003}{9900}\)

Bài 5:

a: \(3,\left(15\right)=3+\dfrac{15}{99}=3+\dfrac{5}{33}=\dfrac{3\cdot33+5}{33}=\dfrac{104}{33}\)

b: \(0,2\left(07\right)=0,2+0,0\left(07\right)=\dfrac{41}{198}\)

c: \(0,1\left(37\right)=0,1+0,0\left(37\right)=\dfrac{1}{10}+\dfrac{37}{990}=\dfrac{68}{495}\)

d: \(0,20\left(23\right)=0,20+0,00\left(23\right)=0,2+\dfrac{23}{9900}=\dfrac{2003}{9900}\)

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