tách bài ra nha b

1 being 

2 to do

3 moving

4 to do

5 take/going

6 leaving 

7 repairing

8 playing

9 to use

10 speaking

`a)` Cho `x-1/2x^2=0`

`=>x(1-1/2x)=0`

`@TH1: x = 0`

`@TH2: 1-1/2x=0=>1/2x=1=>x=2`

Nghiệm của đa thức là `x=0` hoặc `x=2`

__________________________________________

`b)\overline{X}=[6.3+7.6+8x+9.4]/[3+6+x+4]`

  Mà `\overline{X}=7,6`

 `=>[96+8x]/[13+x]=7,6`

 `=>96+8x=7,6(13+x)`

 `=>96+8x=98,8+7,6x`

 `=>8x-7,6x=98,8-96`

 `=>0,4x=2,8`

 `=>x=7`

Vậy `x=7`

cho \(x-\dfrac{1}{2}x^2=0\)

\(=>x\left(1-\dfrac{1}{2}x\right)=0\)

\(=>\left[{}\begin{matrix}x=0\\\dfrac{1}{2}x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1:\dfrac{1}{2}=2\end{matrix}\right.\)

\(Q\left(x\right)=x^4+3x^2+1\)

vì \(\left\{{}\begin{matrix}x^4\ge0\\3x^2\ge0\end{matrix}\right.\)

\(=>x^4+3x^2\ge0\)

mà 1 > 0

\(=>x^4+3x^2+1\ge0\)

hay Q(x ) ko có nghiệm

Bài 6.

a)Công suất ấm: \(P=\dfrac{A}{t}=\dfrac{900\cdot1000}{10\cdot60}=1500W\)

Dòng điện qua ấm: \(I=\dfrac{P}{U}=\dfrac{1500}{220}=\dfrac{75}{11}A\)

Điện trở dây nung: \(R=\dfrac{U}{I}=\dfrac{220}{\dfrac{75}{11}}=\dfrac{484}{15}\Omega\)

b)Điện năng tiêu thụ trong 1 tháng (30 ngày):

\(T=900\cdot1000\cdot30\cdot3600=9,72\cdot10^{10}J=27000kWh\)

Tiền điện phải trả: \(T=27000\cdot1500=40500\left(k.đồng\right)\)

c)Công suất tiêu thụ thực: 

\(P=UI=\dfrac{U^2}{R}=\dfrac{110^2}{\dfrac{484}{15}}=375W\)

Bài 7.

CTM: \(\left(Đ_1ntR_b\right)//Đ_2\)

\(R_1=\dfrac{U_{Đ1}^2}{P_{Đ1}}=\dfrac{10^2}{2}=50\Omega;I_{Đ1đm}=\dfrac{P_{Đ1}}{U_{Đ1}}=\dfrac{2}{10}=0,2A\)

\(R_2=\dfrac{U^2_{Đ2}}{P_{Đ2}}=\dfrac{12^2}{3}=48\Omega;I_{Đ2đm}=\dfrac{P_{Đ2}}{U_{Đ2}}=\dfrac{3}{12}=0,25A\)

Để đèn 1 sáng bình thường \(\Rightarrow I_b=I_{Đ1đm}=0,2A\)

\(R_{Đ1+b}=\dfrac{12}{0,2}=60\Omega\)

\(R_b=60-R_{Đ1}=60-50=10\Omega\)

1)

Cho hỗn hợp vào nước

- lọc phần không tan, ta được $MgO,Cu$

- cô cạn dung dịch thu được $ZnSO_4$

Cho phần không tan vào dd HCl, lọc phần không tan thu được Cu

$MgO + 2HCl \to MgCl_2 + H_2O$

Cho phần dung dịch vào NaOH

$MgCl_2 + 2NaOH \to Mg(OH)_2 + 2NaCl$

Cho nung phần kết tủa thu được MgO
$Mg(OH)_2 \xrightarrow{t^o} MgO + H_2O$

3 C
4 B

5 A

6 B

7 A
8 A

9 C

10 D

11 B

12 B

13 A

14 C

15 B

16 A

17 D

18 D

19 C

20 A

21 B

22 B

23 C

24 D

25 B

26 D

27 A

28 D

29 C

30 B

31 C

32 A

33 A

Viết lại câu

12 

Authorities in Ho Chi Minh City suggested that current school break schould be extended by us until the end of March to control the epidemic caused by Covid-19

15 The plumber who has just been had check the pipes in my house is very skillful

16 In spite of the fact that we had invited her to the party, she didn't come

17 Since they didn't have a map, they lost their way

18 I suggest that energy-saving bulbs should be used to reduce the electricity bill

19 We are looking forward to seeing your gook action from your company 

20 Linda congratulated Owen on scoring a wonderful goal

21 She had all the water pipes in her house checked by the plumber yesterday

22 If you go on smoking, you will be seriously sick

23 I suggested all the bag should be collected and taken to the garbage dump

24 If you aren't carefull, you will fall over the rocks

25 If you work too much, you will feel tired

26 If you are careless, you'll break the vase

27 Because of the fact that I am sick, I don't go out

28 Because of the fact that the water is polluted, we can't swim in that river

29 Because of the fact that Nga has had a sore leg, she had difficulty in walking

30 Because of having behaved badly, he found himself in trouble

31 Because he was seriously sick, he couldn't come to the class

32 Because of too much exhaust fume, people have to wear.........

a) Các góc kề bù nhau là: 

1. \(\widehat{xOy}\) và \(\widehat{xOt}\)

2. \(\widehat{yOz}\) và \(\widehat{zOt}\)    

b) Ta có: \(\widehat{yOt}\) là góc bẹt \(\Rightarrow\widehat{yOt}=180^o\)

Mà \(\widehat{xOy}\) và \(\widehat{xOt}\) kề bù \(\Rightarrow\widehat{xOy}+\widehat{xOt}=\widehat{yOt}\)

\(\Rightarrow\widehat{xOt}=\widehat{yOt}-\widehat{xOy}=180^o-45^o=135^o\) 

Ta có: \(\widehat{xOz}=\widehat{xOy}+\widehat{yOz}=45^o+30^o=75^o\)

Mà \(\widehat{yOz}\) và \(\widehat{zOt}\) kề bù \(\Rightarrow\widehat{yOz}+\widehat{zOt}=\widehat{yOt}=180^o\)

\(\Rightarrow\widehat{zOt}=\widehat{yOt}-\widehat{yOz}=180^o-30^o=150^o\)

8:

b: ΔMBA có MD là phân giác của góc BMA

nên BD/DA=BM/MA

ΔBMC có ME là phân giác

nên BE/EC=BM/MC

mà MA=MC

nên BE/EC=BM/MA=BD/DA

=>DE//AC