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18 tháng 8 2017

I. Write the words with the sounds:

1. school 5. pronunciation

2. clothes 6. bring

3. special 7. climb

4. black board 8. breakfast

II. Put the following words in the correct column according to their parts of speech: cart, usually, against, according to, adore, region, badly, satisfied, among,gather,southern, field,relax, relaxing, finally.

Verbs Adjective Noun Adverb Preposition
adore satisfied cart usually against
gather southern field badly among
relax relaxing region finally according to

III. Complete the words with a,e,i and u:

favorite button envious opportunity

posture forest speciality style

IV. Give the adjectives from the following verbs:

1. to create __creative__

2. to excite __exciting/excited__

3. to develop __developable__

4. to interest __interesting/interested__

5. to forget __forgettable__

6. to bore __boring/bored__

7. to help ___helpable__

8. to adore ___adoring/adored__

24 tháng 6 2021

1 doesn't have - has

2 watches - is watching

3 swim - go

4 play - are playing

5 rides - is walking

6 works

7 belongs

8 sees

9 teaches

10 is sleeping

11 is knocking

12 is coming

13 is having

14 learn

15 smoke

16 collect

17 eat

18 are

19 do your parents do

20 will travel

21 is going to visit

22 will come

23 practice - will speak

24 play

25 is - doing

26 is - is cooking

27 washes

28 come

29 to come

30 to drink

31 like eating

32 likes listening

33 do you like to do

34 ride

35 do

36 stay

37 go

38 sing

39 dancing -

22will be

25 will be

50 playing - 26 am having

51 go

52 isn't

53 is - is

54 Do - camp

55 Do - plant

11 tháng 11 2021

(╯‵●ω●′)╯🍫🍬🍭
Hai bn:>

19 tháng 4 2021

1 visit

2 hotel

3 near

4 travel

5 convenient

6 map

7 clothes

8 camera

9 take

10 attractions

a: Thay x=25/16 vào A, ta được:

\(A=\left(\dfrac{5}{4}+1\right):\left(\dfrac{5}{4}-3\right)=\dfrac{9}{4}:\dfrac{-7}{4}=\dfrac{-9}{7}\)

b: \(B=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\)

\(=\dfrac{-3\sqrt{x}-3}{x-9}\)

a: Xét ΔADB và ΔADC có

AD chung

DB=DC

AB=AC

Do đó: ΔADB=ΔADC

Suy ra: \(\widehat{BAD}=\widehat{CAD}\)

hay AD là tia phân giác của góc BAC

b: ta có: ΔABC cân tại A

mà AD là đường trung tuyến

nên AD là đường cao

12 tháng 3 2022

câu c,d đâu vậy chị ???

20 tháng 1 2019

???

20 tháng 1 2019

Mk cx là thành viên mới

24 tháng 9 2021

\(a,ĐK:x\le\dfrac{1}{5}\\ PT\Leftrightarrow1-5x=9\Leftrightarrow x=-\dfrac{8}{5}\\ b,ĐK:x\ge\dfrac{3}{5}\\ PT\Leftrightarrow\sqrt{5x-3}\left(\sqrt{5x+3}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x-3=0\\\sqrt{5x+3}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\5x+3=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=\dfrac{1}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{3}{5}\)

\(c,ĐK:x\ge0\\ PT\Leftrightarrow2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ge0\\ PT\Leftrightarrow x-4\sqrt{x}+4-3=0\\ \Leftrightarrow\left(\sqrt{x}-2-\sqrt{3}\right)\left(\sqrt{x}-2+\sqrt{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2+\sqrt{3}\\\sqrt{x}=2-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7+4\sqrt{3}\left(tm\right)\\x=7-4\sqrt{3}\left(tm\right)\end{matrix}\right.\\ e,ĐK:x\ge3\\ PT\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\Leftrightarrow x=28\left(tm\right)\)

AH
Akai Haruma
Giáo viên
24 tháng 9 2021

Lời giải:

a. ĐKXĐ: $x\leq \frac{1}{5}$

PT $\Leftrightarrow 1-5x=3^2=9$

$\Leftrightarrow 5x=-8\Leftrightarrow x=\frac{-8}{5}$ (tm)

b. ĐKXĐ: $x\geq \frac{3}{5}$

PT $\Leftrightarrow 25x^2-9=4(5x-3)$

$\Leftrightarrow (5x-3)(5x+3)-4(5x-3)=0$

$\Leftrightarrow (5x-3)(5x-1)=0$

$\Leftrightarrow x=\frac{3}{5}$ (tm) hoặc $x=\frac{1}{5}$ (loại)

c. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow x-4\sqrt{x}+3=0$

$\Leftrightarrow (\sqrt{x}-1)(\sqrt{x}-3)=0$

$\Leftrightarrow \sqrt{x}=1$ hoặc $\sqrt{x}=3$

$\Leftrightarrow x=1$ hoặc $x=9$

d. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow (\sqrt{x}-2)^2-5=0$

$\Leftrightarrow (\sqrt{x}-2)^2=5$
$\Leftrightarrow \sqrt{x}-2=\pm \sqrt{5}$

$\Leftrightarrow \sqrt{x}=2+\sqrt{5}$ (chọn) hoặc $\sqrt{x}=2-\sqrt{5}$ (loại do âm)

$\Leftrightarrow x=(2+\sqrt{5})^2=9+4\sqrt{5}$

e.ĐKXĐ: $x\geq 3$
PT $\Leftrightarrow 2\sqrt{9}.\sqrt{x-3}-\frac{1}{5}.\sqrt{25}.\sqrt{x-3}-\frac{1}{7}\sqrt{49}.\sqrt{x-3}=20$

$\Leftrightarrow 6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20$

$\Leftrightarrow 4\sqrt{x-3}=20$

$\Leftrightarrow \sqrt{x-3}=5$

$\Leftrightarrow x-3=25$

$\Leftrightarrow x=28$