1. goes - go
2. plays
3. studies
4. flies
5. designs
6. eat
7. works
8. grows
9. sings
10. do
11. cook - cooks
12. carries - carry

chữ hơi nhỏ á

1. goes - go

2. plays

3. studies

4. fly

5. designs

6. eat

7. works

8. grows

9. sings

10. do

11. cook - cooks

12. carries - carry

a, 4,78+x=5,6 <=> x=5,6-4,78 = 0,82 

b, x-3,92=0,61 <=> x = 3,92 + 0,61 = 4,53

c, 7-x=4,52 <=> x = 7-4,52 = 2,48 

a: =>x=5,6-4,78=0,82

b: =>x=0,61+3,92=4,53

c: =>x=7-4,52=2,48

d: =>x=17,1:4,5=3,8

e: =>x=17,1x4,5=76,95

f: =>x=42,16:6,2=6,8

1. What has Jim recently experienced ?

2. where do the ethnic minorities in Viet Nam often live ?

3. How are their costumes ?

4. Where do they often gather in special occasions ?

5. What does the chief of the community often tell at the communal house ?

6. When do ethnic people often hold festival ?

Bài 3: 

Xét ΔIAB có 

\(\widehat{AIB}+\widehat{IAB}+\widehat{IBA}=180^0\)

\(\Leftrightarrow\widehat{IAB}+\widehat{IBA}=115^0\)

hay \(\widehat{DAB}+\widehat{ABC}=230^0\)

Xét tứ giác ABCD có 

\(\widehat{D}+\widehat{C}+\widehat{DAB}+\widehat{CBA}=360^0\)

\(\Leftrightarrow\widehat{D}+\widehat{C}=150^0\)

mà \(\widehat{C}-\widehat{D}=10^0\)

nên \(2\cdot\widehat{C}=160^0\)

\(\Leftrightarrow\widehat{C}=80^0\)

\(\Leftrightarrow\widehat{D}=70^0\)

Ta có (2004+1)=2003.2004+2003

(2003+1)=2003.2004+2004

Vì 2003<2004 nên 2003.2004+2003<2003.2004+2004

Vậy 2003.2005<2004^2

Ta có 

Vì 2003<2004 nên 2003.2004+2003<2003.2004+2004

Vậy A<B
tick nha để mk làm câu b

\(Ta có: 8 ( 7 8 + 1 ) ( 7 4 + 1 ) ( 7 2 + 1 ) = 1/ 6 .48 ( 7 2 + 1 ) ( 7 4 + 1 ) ( 7 8 + 1 ) = 1/ 6 ( 7 2 − 1 ) ( 7 2 + 1 ) ( 7 4 + 1 ) ( 7 8 + 1 ) = 1/ 6 ( 7 4 − 1 ) ( 7 4 + 1 ) ( 7 8 + 1 ) = 1/ 6 ( 7 8 − 1 ) ( 7 8 + 1 ) = 1/ 6 ( 7 16 − 1 ) Vì 7 16 − 1 > 1/ 6 ( 7 16 − 1 ) nên 7 16 − 1 > 8 ( 7 8 + 1 ) ( 7 4 + 1 ) ( 7 2 + 1 )\)

a, \(\left\{{}\begin{matrix}35x-28y=21\\35x-45y=40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17y=-19\\x=\dfrac{3+4y}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{19}{17}\\x=-\dfrac{13}{17}\end{matrix}\right.\)

b, Đặt x;y khác 0 

Đặt 1/x = t ; 1/y = u 

\(\left\{{}\begin{matrix}t-8u=18\\5t+4u=51\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5t-40u=90\\5t+4u=51\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-44u=39\\t=18+8u\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u=-\dfrac{39}{44}\\t=\dfrac{120}{11}\end{matrix}\right.\)

Theo cách đặt y = -44/39 ; x = 11/120 (tm)

\(a,\\ \Leftrightarrow\left\{{}\begin{matrix}35x-28y=21_{\left(1\right)}\\35x-45y=40_{\left(2\right)}\end{matrix}\right.\\ Lấy\left(1\right)-\left(2\right),ta.đc:\\ -17y=19\Leftrightarrow y=\dfrac{-19}{17}\\ Thay.vào.\left(1\right):\\ 35x-28.\dfrac{-19}{17}=21\Leftrightarrow x=\dfrac{-5}{17}\) 

Vậy ......

\(b,\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=18+\dfrac{1}{y}\\5.\left(18+\dfrac{1}{y}\right)+\dfrac{4}{y}=51\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=18+\dfrac{1}{y}\\\dfrac{9}{y}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=18-\dfrac{13}{3}\\\dfrac{1}{y}=\dfrac{-13}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{41}{3}\\\dfrac{1}{y}=\dfrac{-13}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{41}\\y=\dfrac{-3}{13}\end{matrix}\right.\)

b: Xét ΔBID có \(\widehat{DBI}=\widehat{DIB}\left(=\widehat{IBC}\right)\)

nên ΔBID cân tại D

Xét ΔEIC có \(\widehat{EIC}=\widehat{ECI}\left(=\widehat{ICB}\right)\)

nên ΔEIC cân tại E

c: Ta có: DE=DI+IE

mà DI=DB

và EC=IE

nên DE=DB+EC