\(\left(1\right)FeCl_2+2NaOH\to Fe\left(OH\right)_2\downarrow +2NaCl\\ \left(2\right)Fe\left(OH\right)_2\xrightarrow{t^0}FeO+H_2O\\ \left(3\right)FeO+C\to Fe+CO\uparrow \\ \left(4\right)2Fe+3Cl_2\xrightarrow[]{t^o}2FeCl_3\\ \left(5\right)FeCl_3+3NaOH\to Fe\left(OH\right)_3\downarrow +3NaCl\\ \left(6\right)2Fe\left(OH\right)_3\xrightarrow{t^o}Fe_2O_3+3H_2O\\ \left(7\right)Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2\\ \left(8\right)Fe+2HCl\to FeCl_2+H_2\)

I/

2. visited

3. am

4. will lend

5. would get

6. came

7. will see

8. gets

9. would tell

10. will wash

11. won't see

12. ate

13. have

14. got

15. will buy

II/

2. tell/will help

3. wouldn't marry/was

4. didn't fall/ wouldn't break

5. sings/will join

6. would go/were

7. was/would go

8. saw/would say

9. would go/ wasn't 

10. had/would give

11. played/would won

12. were/would accept

13. would have/didn't rain

14. went/would see

15. were/would help

16. knew/would go

18. doesn't lend/won't be

a: R1//R2

=>I=I1+I2=1,6(A)

b: R tđ=11,2/1,6=7(\(\Omega\))

\(a)I=I_1+I_2=0,5+1,1=1,6A\\ b)R_{tđ}=\dfrac{U}{I}=\dfrac{11,2}{1,6}=7\Omega\)

(x + 3)/3 = x/(-12)

-12.(x + 3) = 3x

-12x - 36 = 3x

-12x - 3x = 36

-15x = 36

x = -36/15

x = -12/5

a: Xét ΔADI và ΔAEI có

AD=AE

\(\widehat{DAI}=\widehat{EAI}\)

AI chung

DO đó: ΔADI=ΔAEI

Suy ra: DI=EI

hay I là trung điểm của DE

b: Xét ΔNID vuông tại I và ΔNIE vuông tại I có 

NI chung

ID=IE

Do đó: ΔNID=ΔNIE

Suy ra: ND=NE

So sánh đúng không á.

\(\sqrt{4}\left(1+\sqrt{5}\right)\\ =2\left(1+\sqrt{5}\right)=2+\sqrt{10}\)

Mà 10>5 nên \(\sqrt{10}>\sqrt{5}\)

=> \(2+\sqrt{5}< \sqrt{4}\left(1+\sqrt{5}\right)\)