Giúp em đi đc ko ạ

III. Use the correct tense or form of the verb in parentheses:

1 taught

2 enjoys

3 living

4 to carry

5 died

6 to choose

7to accept

8 came

9 could / played / was

10 goes

11 could not

12 tidy

13 visited / driving / to driving / are / to transport

14 is learning / to drive / had / took / was / drove / felt / finished / went / felt

IV. Use the correct form of the word in parentheses:

1 magician / interested / magical

2 electricity

3 equipment

4 electric / electrical

5 traditional

6 Unfortunately

7 decision

8 appeared / magically

9 wisdom

10 excitedly

11 electrify

12 broken

13 celebrations

14 comfortable

15 immediately

16 fortunate

17 traditionally

18 cruelty

19 poverty

20 amazement

21 marriage

22 servant

23 imagination

24 disappearance

25 modernize

Bài 2:

a. 3x(x - 6) - 2x² = x² + 6

<=> 3x² - 18x - 2x² - x² - 6 = 0

<=> 3x² - 2x² - x² - 18x - 6 = 0

<=> -18x - 6 = 0

<=> -18x = 6

<=> x = \(\dfrac{6}{-18}=\dfrac{-1}{3}\)

b. (x - 3)(x - 2) - 5 = x² - 4x

<=> x² - 2x - 3x + 6 - 5 - x² + 4x = 0

<=> x² - x² - 2x - 3x + 4x + 6 - 5 = 0

<=> -x + 1 = 0

<=> -x = -1

<=> x = 1

c. (x + 5)² - 8x = x² + 15

<=> x² + 10x + 25 - 8x - x² - 15 = 0

<=> x² - x² + 10x - 8x + 25 - 15 = 0

<=> 2x + 10 = 0

<=> 2x = -10

<=> x = -5

d. x² - 4x + 4 = 0

<=> x² - 2.2.x + 2² = 0

<=> (x - 2)² = 0

<=> x - 2 = 0

<=> x = 2

e. x² + 8x + 16 = 0

<=> x² + 2.x.4 + 4² = 0

<=> (x + 4)² = 0

<=> x + 4 = 0

<=> x = -4

f. x² - 36 = 0

<=> x² - 6² = 0

<=> (x - 6)(x + 6) = 0

<=> \(\left[{}\begin{matrix}x-6-0\\x+6=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

g. (x + 3)² - 16 = 0

<=> (x + 3)² - 4² = 0

<=> (x + 3 + 4)(x + 3 - 4) = 0

<=> (x + 7)(x - 1) = 0

<=> \(\left[{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\)

k: Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-2x^3+8\)

\(=x^3-8-2x^3+8\)

\(=-x^3\)

Đề bài thiếu thì phải

32, A

33,B

34,C

Gì vậy? Bài đâu?

a) Xét (O) có

\(\widehat{BAD}\) là góc nội tiếp chắn \(\stackrel\frown{BD}\)

\(\widehat{CAD}\) là góc nội tiếp chắn \(\stackrel\frown{CD}\)

mà \(\widehat{BAD}=\widehat{CAD}\)(AD là tia phân giác của \(\widehat{BAC}\))

nên \(\stackrel\frown{BD}=\stackrel\frown{CD}\)

hay BD=CD

Ta có: OB=OC(=R)

nên O nằm trên đường trung trực của BC(Tính chất đường trung trực của một đoạn thẳng)(1)

Ta có: BD=CD(cmt)

nên D nằm trên đường trung trực của BC(Tính chất đường trung trực của một đoạn thẳng)(2)

Từ (1) và (2) suy ra OD là đường trung trực của BC

hay OD\(\perp\)BC(đpcm)

1)since

2)for,never

3)for,since

4)since,never

5)for,never

6)since,ever

7)for,since

8)since,ever,never

9)for,since

10)for,never,ever

\(\dfrac{2\left(5x+2\right)}{9}-1=\dfrac{4\left(33+2x\right)}{5}-\dfrac{5\left(1-11x\right)}{9}\)

\(\dfrac{10\left(5x+2\right)}{45}-\dfrac{45}{45}=\dfrac{36\left(33+2x\right)}{45}-\dfrac{25\left(1-11x\right)}{45}\)

\(50x-20-45=1188+72x-25+275x\)

\(50x-25=347x+1163\)

\(50x-347x=25+1163\)

\(-297x=1188\)

\(x=4\\ \)

d) 

\(\dfrac{2\left(x-4\right)}{3}+\dfrac{3x+13}{8}=\dfrac{2\left(2x-3\right)}{5}+12\)

\(\dfrac{80\left(x-4\right)}{120}+\dfrac{15\left(3x+13\right)}{120}=\dfrac{40\left(2x-3\right)}{120}+\dfrac{1440}{120}\)

\(80x-320+45x+195=80x-120+1440\)

\(125x-125=80x+1320\)

\(125x-80x=125+1320\)

\(45x=1445\)

   \(x=\dfrac{1445}{45}\) \(=\dfrac{289}{9}\)

\(a,a^2-10a+25=\left(a-5\right)^2\\ b,4x^2+4x+1=\left(2x+1\right)^2\\ c,4x^2-9=\left(2x-3\right)\left(2x+3\right)\\ d,x^3+3x^2+3x+1=\left(x+1\right)^3\\ e,a^3-3a^2b+3ab^2-b^3=\left(a-b\right)^3\\ f,y^3+8=\left(y+2\right)\left(y^2-2y+4\right)\\ g,27x^3-1=\left(3x-1\right)\left(9x^2+3x+1\right)\)

\(a^2-10x+25=\left(a-5\right)^2\)

b/ \(4x^2+4x+1=\left(2x+1\right)^2\)

c/ \(4b^2-9=\left(2b-3\right)\left(2b+3\right)\)

d/ \(x^3+3x^2+3x+1=\left(x+1\right)^3\)

e/ \(a^3-3a^2b+3ab^2-b^3=\left(a-b\right)^3\)

f/ \(y^3+8=\left(y+2\right)\left(y^2-2y+4\right)\)

g/ \(27x^3-1=\left(3x-1\right)\left(9x^2+3x+1\right)\)