khó nhìn quá

Bài 10:

1: \(\left(7-\frac15+\frac13\right)-\left(6+\frac95+\frac43\right)\)

\(=7-\frac15+\frac13-6-\frac95-\frac43\)

\(=\left(7-6\right)+\left(-\frac15-\frac95\right)+\left(\frac13-\frac43\right)\)

=1-2-1

=-2

2: \(7+\left(\frac{7}{12}-\frac12+3\right)-\left(\frac{1}{12}+5\right)\)

\(=7+\frac{1}{12}+3-\frac{1}{12}-5\)

=10-5

=5

3: \(\left(\frac12-\frac13\right)-\left(\frac53-\frac32\right)+\left(\frac73-\frac52\right)\)

\(=\frac12-\frac13-\frac53+\frac32+\frac73-\frac52\)

\(=-\frac12+\frac13=\frac{-3+2}{6}=-\frac16\)

4: \(\left(\frac27-\frac94\right)-\left(-\frac37+\frac54\right)-\left(\frac24-\frac97\right)\)

\(=\frac27-\frac94+\frac37-\frac54-\frac24+\frac97\)

\(=\left(\frac27+\frac37+\frac97\right)+\left(-\frac94-\frac54-\frac24\right)=\frac{14}{7}-\frac{16}{4}=2-4=-2\)

5: \(\left(\frac53-\frac37+9\right)-\left(2+\frac57-\frac23\right)+\left(\frac87-\frac43-10\right)\)

\(=\frac53-\frac37+9-2-\frac57+\frac23+\frac87-\frac43-10\)

\(=\left(\frac53+\frac23-\frac43\right)+\left(-\frac37-\frac57+\frac87\right)+\left(9-2-10\right)\)

\(=\frac33+\left(-3\right)=1-3=-2\)

Bài 11:

1: \(\frac25\cdot\frac38_{}+\frac58\cdot\frac25=\frac25\left(\frac38+\frac58\right)=\frac25\cdot\frac88=\frac25\)

2: \(\frac23\cdot\frac52-\frac34\cdot\frac23=\frac23\left(\frac52-\frac34\right)=\frac23\cdot\frac74=\frac{14}{12}=\frac76\)

3: \(\frac57\cdot\frac{19}{23}-\frac{12}{23}\cdot\frac57=\frac57\left(\frac{19}{23}-\frac{12}{23}\right)=\frac57\cdot\frac{7}{23}=\frac{5}{23}\)

4: \(\frac72\cdot\frac{11}{6}-\frac72\cdot\frac56=\frac72\left(\frac{11}{6}-\frac56\right)=\frac72\cdot\frac66=\frac72\)

5: \(\frac{11}{9}\cdot\frac34-\frac29\cdot\frac34=\frac34\left(\frac{11}{9}-\frac29\right)=\frac34\cdot\frac99=\frac34\)

6: \(\frac37\cdot\frac{13}{5}+\frac37\cdot\frac85=\frac37\left(\frac{13}{5}+\frac85\right)=\frac37\cdot\frac{21}{5}=\frac{21}{7}\cdot\frac35=3\cdot\frac35=\frac95\)

7: \(\frac{7}{15}\cdot\frac{16}{13}+\frac{7}{15}\cdot\frac{-3}{13}=\frac{7}{15}\left(\frac{16}{13}-\frac{3}{13}\right)=\frac{7}{15}\cdot\frac{13}{13}=\frac{7}{15}\)

8: \(-\frac{23}{7}\cdot\frac{3}{10}+\frac{13}{7}\cdot\frac{3}{10}=\frac{3}{10}\left(-\frac{23}{7}+\frac{13}{7}\right)=\frac{3}{10}\cdot\frac{-10}{7}=-\frac37\)

9: \(\frac{-11}{8}\cdot\frac{19}{3}+\frac{19}{3}\cdot\frac{-5}{8}=\frac{19}{3}\left(-\frac{11}{8}-\frac58\right)=\frac{19}{3}\cdot\left(-2\right)=-\frac{38}{3}\)

Bài 12: Bài 12:

1: \(\frac{-5}{17}\cdot\frac{31}{33}+\frac{-5}{17}\cdot\frac{2}{33}+1\frac{5}{17}\)

\(=-\frac{5}{17}\cdot\left(\frac{31}{33}+\frac{2}{33}\right)+1+\frac{5}{17}\)

\(=-\frac{5}{17}+1+\frac{5}{17}=1\)

2: \(\frac57\cdot\left(-\frac{3}{11}\right)+\frac57\cdot\left(-\frac{8}{11}\right)+2\frac57\)

\(=-\frac57\left(\frac{3}{11}+\frac{8}{11}\right)+2+\frac57\)

\(=-\frac57+2+\frac57=2\)

3: \(\frac{9}{10}\cdot\frac{23}{11}-\frac{1}{11}\cdot\frac{9}{10}+\frac{9}{10}\)

\(=\frac{9}{10}\left(\frac{23}{11}-\frac{1}{11}+1\right)\)

\(=\frac{9}{10}\cdot\left(2+1\right)=\frac{9}{10}\cdot3=\frac{27}{10}\)

4: \(\frac54\cdot\frac{8}{15}+\frac{-5}{16}\cdot\frac{8}{15}-1\)

\(=\frac{8}{15}\left(\frac54-\frac{5}{16}\right)-1\)

\(=\frac{8}{15}\left(\frac{20}{16}-\frac{5}{16}\right)-1=\frac{8}{16}-1=-\frac{8}{16}=-\frac12\)

5: \(-\frac{19}{3}\cdot\frac{14}{4}+\frac{25}{4}\cdot\frac{-19}{3}+4\frac34\)

\(=-\frac{19}{4}\left(\frac{14}{3}+\frac{25}{3}\right)+4\frac34\)

\(=-\frac{19}{4}\cdot13+\frac{19}{4}=\frac{19}{4}\left(-13+1\right)=\frac{19}{4}\cdot\left(-12\right)=-57\)

6: \(\frac{1}{27}\cdot\frac{-3}{7}-\frac59\cdot\frac{-3}{7}+\frac19\)

\(=-\frac37\left(\frac{1}{27}-\frac59\right)+\frac19\)

\(=-\frac37\left(\frac{1}{27}-\frac{15}{27}\right)+\frac19=-\frac37\cdot\frac{-14}{27}+\frac19=\frac29+\frac19=\frac39=\frac13\) b

a/

\(\widehat{BCE}=\widehat{CED}=30^o\)

Hai góc trên ở vị trí sole trong => BC//DE

b/

Ta có

BC//DE (cmt) \(\Rightarrow\widehat{AFB}=180^o-\widehat{EDF}\) (Hai góc trong cùng phía bù nhau)

\(\Rightarrow\widehat{AFB}=180^o-135^o=45^o\)

a, B C A D E

Ta có: góc aCd = góc CDb' (=60°)

Mà 2 góc này ở vị trí so le trong

=> aa' // bb'

Ta có: góc aCd = góc CDb' (=60°)

Mà 2 góc này ở vị trí so le trong

=> aa' // bb'

Xét ∆ABC và ∆DBC có: 

AB = BD 

Góc ABC = góc CBD 

Góc BAC = góc BDC 

=> ∆ABC = ∆DBC

xét ΔABC và ΔDBC, ta có :

góc A = góc D (gt)

BC là cạnh chung

góc ABC = góc DBC

=> ΔABC = ΔDBC, (g.c.g)

Ta có:

\(a< b,c< d,m< n\)

\(\Rightarrow a+c+m< b+d+n\Rightarrow2a+2c+2m< a+b+c+d+m+n\)

\(\Rightarrow a+c+m< \frac{1}{2}\left(a+b+c+d+m+n\right)\)

\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\) ( đpcm )

giúp mình với :V

ta có góc QNO = 180⁰ - 45⁰ = 135⁰

vì tổng 4 góc của một tứ giác bằng 360⁰

xét tứ giác QNOP 

 ta có góc NOP + góc QPO =  360⁰ - ( 45⁰ + 135⁰) = 180⁰ (đpcm)

#\(N\)

`a,` Xét Tam giác `MPH` và Tam giác `MQH` có:

`MP = MQ (g``t)`

`MH` chung

\(\widehat{MHP}=\widehat{MHQ}=90^0\)

`=>` Tam giác `MPH =` Tam giác `MQH (ch - cgv)`

`=>`\(\widehat{MPH}=\widehat{MQH}\) `( 2` góc tương ứng `)`

`b,` Vì Tam giác `MPH =` Tam giác `MQH (a)`

`=>` \(\widehat{PMH}=\widehat{QMH}\) `( 2` góc tương ứng `)`

`=> MH` là tia phân giác của \(\widehat{PMQ}\) 

`c,` Ta có: \(\widehat{MPH}=\widehat{MQH}=50^0\) `(CMT)`

Xét Tam giác `MQH` có:

\(\widehat{MHQ}+\widehat{MQH}+\widehat{QMH}=180^0\) `(`đlí tổng `3` góc trong `1` tam giác `)`

\(90^0+50^0+\widehat{QMH}=180^0\)

`->`\(\widehat{QMH}=180^0-90^0-50^0=40^0\)