\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)

B=2+22+23+24+...+299+2100=2(1+22+23+24)+...+296(1+22+23+24)=2.31+26.31+...+296.31=31(2+26+...+296)⋮31

Bài 1: 

a: Xét ΔABI và ΔACI có

AB=AC

AI chung

BI=CI

Do đó: ΔABI=ΔACI

I

1 will have

2 would have

3 have

4 had

5 will make

6 would make

7 has

8 had

9 is

10 were

II

1 have

2 had

3 had had

4 will go

5 would go

6 would have gone

7 is

8 were - would visit

9 had been - would have visited

10 wouldn't be - were

I.

1. will have

2. would have

3. have

4. had

5. will made

6. would made

7. has

8. had

9. is

10. were

II.

1. have

2. had

3. had had

4. will go

5. would go

6. would have gone

7. is

8. were - would visit

9. had been - would have visited

10. will not be - is

Khó đấy 

Bài 1: 

Vì (d)//y=-2x+1 nên a=-2

Vậy: y=-2x+b

Thay x=1 và y=2 vào (d),ta được:

b-2=2

hay b=4

lỗi

lỗi

Lỗi rồi bạn ơi

\(\dfrac{4}{7}+\dfrac{2}{9}+\dfrac{1}{4}+\dfrac{3}{7}+\dfrac{7}{9}+\dfrac{75}{100}\\ =\dfrac{4}{7}+\dfrac{2}{9}+\dfrac{1}{4}+\dfrac{3}{7}+\dfrac{7}{9}+\dfrac{3}{4}\\ =\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\left(\dfrac{2}{9}+\dfrac{7}{9}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =\dfrac{7}{7}+\dfrac{9}{9}+\dfrac{4}{4}\\ =1+1+1\\ =3\)

Ta có: \(\dfrac{7}{11}=\dfrac{7\times3}{11\times3}=\dfrac{21}{33};\dfrac{8}{11}=\dfrac{8\times3}{11\times3}=\dfrac{24}{33}\)

2 phân số giữa là: \(\dfrac{22}{33};\dfrac{23}{33}\)

\(=\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\left(\dfrac{2}{9}+\dfrac{7}{9}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)=1+1+1=3\)

(\(\dfrac{4}{7}\)+\(\dfrac{3}{7}\))+(\(\dfrac{2}{9}\)+\(\dfrac{7}{9}\))+(\(\dfrac{1}{4}\)+\(\dfrac{75}{100}\))

=1+1+1

=3