Bài 2: 

b: f(0)=5

f(-1)=8

f(-3)=14

thank bn

Đáp án không thôi nhá chứ giải mất tg lắm

Vg cậu 

1 are

2 am

3 is

4 are

5 are

6 are

7 is

8 is

9 is

10 are

IV

1 is writing

2 are losing

3 is having

4 is staying

5 am not lying

6 is always using

7 are having

8 Are you playing

9 are not touching

10 Is - listening

11 Is- winning

12 am not staying

13 is not working

14 is not reading

15 isn't raining

16 am not listening

17 Are they making

18 Are you doing

19 Is - sitting

20 is - doing

21 are-putting

22 are-wearing

23 is-studying

2, am

3, is

4,are

5,are

6,are

7,is

8,is

9,is

10,are

IV

1,2,7 OK

3,is having

4,has stayed

5,am not lying

6,always uses

8,Are-playing

9,not to touch

10,Is-listening

11,Are-winning

12,am not staying

13,isn't working

14,isn't reading

15,isn't raining

16,am not listening

17,Are-making

18,Are-doing

19,Is-sitting

20,is-doing

21,do-putting

22,do-wear

23,is-studying

27. unforgetable
câu 31 đổi thành ...if I can give him his guide....
câu 32 thiếu on trong turned on và đổi thành has been turned on vì câu gốc là HTHT
câu 33 rút gọn mệnh đề ok
câu 34 hơi phân vân nhưng mình thấy đúng

beautifully

excitedly

interesting

collected

arrangements

peacefully

excited

different

Cảm ơn cậu rất nhiều ạ 

a,\(x^2-7x+6=x^2-x-6x+6\)

                       \(=x\left(x-1\right)-6\left(x-1\right)\)

                        \(=\left(x-6\right)\left(x-1\right)\)

a) x²-7x+6=(x²-x)-(6x-6)=x(x-1)-6(x-1)=(x-6)(x-1)

b) x²-6x+3=(x²-6x+9)-6=(x-3)²-\(\sqrt{6^2}\)=(x-3-\(\sqrt{6}\))(x-3+\(\sqrt{6}\))

c) x²-4x+3=(x²-x)-(3x-3)=x(x-1)-3(x-1)=(x-3)(x-1)

d) 3x²-5x+2=(3x²-3x)-(2x-2)=3x(x-1)-2(x-1)=(3x-2)(x-1)

e) 7x²-x-6=(7x²-7x)+(6x-6)=7x(x-1)+6(x-1)=(7x+6)(x-1)

f) 3x²-5x-8=(3x²+3x)-(8x+8)=3x(x+1)-8(x+1)=(3x-8)(x+1)

g) x²-6x+5=(x²-x)-(5x-5)=x(x-1)-5(x-1)=(x-5)(x-1)

h) x²-2x-3=(x²-2x+1)-4=(x-1)²-2²=(x-1-2)(x-1+2)=(x-3)(x+1)

i) x²-x-12=(x²+3x)-(4x+12)=x(x+3)-4(x+3)=(x-4)(x+3)

a: Xét (O) có 

ΔAMB nội tiếp đường tròn

AB là đường kính

Do đó: ΔAMB vuông tại M

Xét tứ giác AMCK có 

\(\widehat{AKC}+\widehat{AMC}=180^0\)

nên AMCK là tứ giác nội tiếp

hay A,M,C,K cùng thuộc một đường tròn

Bài 3. 

a)\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{30\cdot18}{30+18}=11,25\Omega\)

\(R_{tđ}=R_1+R_{23}=20+11,25=31,25\Omega\)

\(I_3=0,8A\Rightarrow U_3=0,8\cdot18=14,4V\)

b)\(\Rightarrow I_m=I_1=I_{23}=\dfrac{U_{23}}{R_{23}}=\dfrac{14,4}{11,25}=1,28A\)

\(U_m=1,28\cdot31,25=40V\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{U_3}{R_2}=\dfrac{14,4}{30}=0,48A\)

Bài 4:

a. Mắc nối tiếp.

 \(U_b=U-U_d=20-12=6V\)

\(I=I_d=I_b=\dfrac{P_d}{U_d}=\dfrac{6}{12}=0,5A\left(R_dntR_b\right)\)

\(\Rightarrow R_b=U_b:I_b=6:0,5=12\Omega\)

b. \(R=p\dfrac{l}{S}\Rightarrow l=\dfrac{R.S}{p}=\dfrac{20.0,05.10^{-6}}{4.10^{-7}}=2,5\left(m\right)\)