a: Vì \(\widehat{xOy};\widehat{zOy}\) là hai góc kề nhau nên tia Oy nằm giữa hai tia Ox,Oz

=>\(\widehat{xOy}+\widehat{zOy}=\widehat{xOz}=120^0\)

mà \(\widehat{xOy}-\widehat{zOy}=40^0\)

nên \(\widehat{xOy}=\dfrac{120^0+40^0}{2}=80^0;\widehat{zOy}=80^0-40^0=40^0\)

b: Ta có: \(\widehat{xOm}+\widehat{xOy}=180^0\)(hai góc kề bù)

=>\(\widehat{xOm}+80^0=180^0\)

=>\(\widehat{xOm}=100^0\)

c: Ta có: \(\widehat{mOn}=\widehat{yOz}\)(hai góc đối đỉnh)

mà \(\widehat{yOz}=40^0\)

nên \(\widehat{mOn}=40^0\)

a) Ta có:

\(\widehat{xOy}+\widehat{zOy}=\widehat{xOz}=120^o\) (hai góc kề nhau)

Mà \(\widehat{xOy}-\widehat{zOy}=40^o\) nên:

\(\widehat{xOy}=\dfrac{120^o+40^o}{2}=80^o\)

Do đó: \(\widehat{zOy}=120^o-80^o=40^o\)

Vậy...

b) Ta có:

\(\widehat{xOy}+\widehat{xOm}=180^o\) (hai góc kề bù)

Mà \(\widehat{xOy}=80^o\) nên:

\(\widehat{xOm}=180^o-80^o=100^o\)

Vậy...

c) Ta có:

\(\widehat{mOn}=\widehat{zOy}\) (hai góc đối đỉnh)

Mà \(\widehat{zOy}=40^o\) nên:

\(\widehat{mOn}=40^o\)

Vậy...

a: \(\dfrac{1}{2}< \dfrac{12}{a}< \dfrac{4}{3}\)

=>\(\dfrac{12}{24}< \dfrac{12}{a}< \dfrac{12}{9}\)

=>9<a<24

mà a nguyên

nên \(a\in\left\{10;11;...;23\right\}\)

b: \(\dfrac{7}{4}< \dfrac{a}{8}< 3\)

=>\(\dfrac{14}{8}< \dfrac{a}{8}< \dfrac{24}{8}\)

=>14<a<24

mà a nguyên

nên \(a\in\left\{15;16;...;23\right\}\)

c: \(\dfrac{2}{3}< \dfrac{a-1}{6}< \dfrac{8}{9}\)

=>\(\dfrac{12}{18}< \dfrac{3\left(a-1\right)}{18}< \dfrac{16}{18}\)

=>12<3a-3<16

=>15<3a<19

=>5<a<19/3

mà a nguyên

nên a=6

d: \(\dfrac{12}{9}< \dfrac{4}{a}< \dfrac{8}{3}\)

=>\(\dfrac{8}{6}< \dfrac{8}{2a}< \dfrac{8}{3}\)

=>3<2a<6

mà a nguyên

nên 2a=4

=>a=2

\(\dfrac{1}{21}\) = \(\dfrac{1\times3}{21\times3}\) = \(\dfrac{3}{63}\) < \(\dfrac{3}{27}\)

Vậy \(\dfrac{1}{21}\) < \(\dfrac{3}{27}\)

a;  \(\dfrac{4}{27}\) = \(\dfrac{4\times7}{27\times7}\) = \(\dfrac{28}{189}\)

     \(\dfrac{15}{63}\) = \(\dfrac{15\times3}{63\times3}\) = \(\dfrac{45}{189}\)

     \(\dfrac{28}{189}\) < \(\dfrac{45}{189}\)

     - \(\dfrac{28}{189}\) > - \(\dfrac{45}{189}\)

Vậy - \(\dfrac{4}{27}\) > \(\dfrac{15}{-63}\) 

b;   \(\dfrac{13}{15}\) = \(\dfrac{13\times9}{15\times9}\) =  \(\dfrac{117}{135}\)

     \(\dfrac{9}{11}\) = \(\dfrac{9\times13}{11\times13}\) = \(\dfrac{117}{143}\)

     \(\dfrac{117}{135}\) > \(\dfrac{117}{143}\) 

Vậy \(\dfrac{13}{15}\) > \(\dfrac{9}{11}\) 

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\\\Leftrightarrow \frac{x+4}{2000}+\frac{x+3}{2001}-\frac{x+2}{2002}-\frac{x+1}{2003}=0\\\Leftrightarrow \left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)-\left(\frac{x+2}{2002}+1\right)-\left(\frac{x+1}{2003}+1\right)=0\\\Leftrightarrow \frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\\\Leftrightarrow (x+2024)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\\\Leftrightarrow x+2024=0(\text{vì }\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003} \ne0)\\\Leftrightarrow x=-2024\)

Vậy phương trình có 1 nghiệm duy nhất là $x=-2024$.

a: \(-\dfrac{19}{49}=\dfrac{-19\cdot47}{49\cdot47}=\dfrac{-893}{2303}\)

\(\dfrac{-23}{47}=\dfrac{-23\cdot49}{47\cdot49}=\dfrac{-1127}{2303}\)

mà -893>-1127

nên \(-\dfrac{19}{49}>-\dfrac{23}{47}\)

b: \(\dfrac{-5}{8}=\dfrac{-5\cdot5}{8\cdot5}=\dfrac{-25}{40}\)

\(\dfrac{7}{-10}=\dfrac{-7}{10}=\dfrac{-7\cdot4}{10\cdot4}=\dfrac{-28}{40}\)

mà -25>-28

nên \(-\dfrac{5}{8}>\dfrac{7}{-10}\)

c: \(\dfrac{24}{35}=\dfrac{24\cdot6}{35\cdot6}=\dfrac{144}{210};\dfrac{19}{30}=\dfrac{19\cdot7}{30\cdot7}=\dfrac{133}{210}\)

mà 144>133

nên \(\dfrac{24}{35}>\dfrac{19}{30}\)

\(-1,25=-\dfrac{125}{100}=-\dfrac{5}{4}\)

a: \(-\dfrac{15}{21}=\dfrac{-15:3}{21:3}=\dfrac{-5}{7}=\dfrac{-5\cdot11}{7\cdot11}=\dfrac{-55}{77}\)

\(\dfrac{-36}{44}=\dfrac{-36:4}{44:4}=\dfrac{-9}{11}=\dfrac{-9\cdot7}{11\cdot7}=\dfrac{-63}{77}\)

mà -55>-63

nên \(-\dfrac{15}{21}>-\dfrac{36}{44}\)

b: \(-\dfrac{16}{30}=\dfrac{-16:2}{30:2}=\dfrac{-8}{15}=\dfrac{-8\cdot4}{15\cdot4}=\dfrac{-32}{60}\)

\(\dfrac{-35}{84}=\dfrac{-35:7}{84:7}=\dfrac{-5}{12}=\dfrac{-5\cdot5}{12\cdot5}=\dfrac{-25}{60}\)

mà -32<-25

nên \(-\dfrac{16}{30}< -\dfrac{35}{84}\)

c: \(\dfrac{-5}{91}=\dfrac{-5\cdot101}{91\cdot101}=\dfrac{-505}{9191}\)

mà -505<-501

nên \(-\dfrac{5}{91}< -\dfrac{501}{9191}\)

a) Rút gọn:

\(-\dfrac{15}{21}=-\dfrac{15:3}{21:3}=-\dfrac{5}{7}\)

\(-\dfrac{36}{44}=-\dfrac{36:4}{44:4}=-\dfrac{9}{11}\)

Quy đồng(MSC:77)

\(-\dfrac{5}{7}=-\dfrac{5.11}{7.11}=-\dfrac{55}{77}\\ -\dfrac{9}{11}=-\dfrac{9.7}{11.7}=-\dfrac{63}{77}\)

Nhận thấy: \(-\dfrac{55}{77}>-\dfrac{63}{77}\Rightarrow-\dfrac{15}{21}>-\dfrac{36}{44}\)

b) Rút gọn:

\(-\dfrac{16}{30}=-\dfrac{16:2}{30:2}=-\dfrac{8}{15}\\ -\dfrac{35}{84}=-\dfrac{35:7}{84:7}=-\dfrac{5}{12}\)

Quy đồng (MSC:60)

\(-\dfrac{8}{15}=-\dfrac{8.4}{15.4}=-\dfrac{32}{60}\\ -\dfrac{5}{12}=-\dfrac{5.5}{12.5}=-\dfrac{25}{60}\)

Nhận thấy: \(-\dfrac{32}{60}< -\dfrac{25}{60}\Rightarrow-\dfrac{16}{30}< -\dfrac{35}{84}\)

c) \(-\dfrac{5}{91}=-\dfrac{5.101}{91.101}=-\dfrac{505}{9191}< -\dfrac{501}{9191}\)