1B

2C

3A

4D

5C

6A

Chào em, em tham khảo nhé:

1. B

2. C

3. A

4. D

5. C

6. A

Chúc em học tốt.

1B

2C

3A

4D

5C

6A

1. shouldn't

2. should

3. should

4. should

5. should

6. Should

7. should

8. shouldn't

9. should

10. should

11. shouldn't

12. Should

13. should

14. shouldn't

15. shouldn't

-----------

II

1. was -> were

2. Was -> Were

3. was -> did

4. went -> go

5. weren't -> didn't

6. have -> had

7. broke -> break

8. go -> went

9. am -> was

10. turn -> turned

--------------

III

1. taught/ to ride/ was

2. does your father go/ drives

3. are going

4. used/ lived 

5. spent/ repairing 

-------------

IV

1. Lang Lieu couldn't buy any special food because he was very pool

2. During Tet, Vietnamese people buy all kinds of sweets, so they make Chung cakes as well

3. The Hung King Temple Festival was a local festival. However, it has become a public holiday in Viet Nam

4. At the Mid-Autumn festival, kids can sing, dance and enjoy moon cakes; therefore, every child like it very much.

5. Because the water is highly polluted, we can't swim in this part of the river

6. Mr. Minh is admired since he dedicates all his life to protecting the environment

-------------

V

1. A

2. B

3. A

4. D

5. C 

Pleaseeeeeee!

Bài 1:

a) \(\dfrac{4}{5}-\dfrac{7}{6}+\dfrac{-6}{15}=\dfrac{24}{30}-\dfrac{35}{30}+\dfrac{-12}{30}=\dfrac{24-35+-12}{30}=\dfrac{-23}{30}\) 

b) \(\dfrac{-5}{9}.\dfrac{7}{13}+\dfrac{6}{13}.\dfrac{-5}{9}+3\dfrac{7}{9}\) 

\(=\dfrac{-5}{9}.\left(\dfrac{7}{13}+\dfrac{6}{13}\right)+\dfrac{34}{9}\) 

\(=\dfrac{-5}{9}.1+\dfrac{34}{9}\) 

\(=\dfrac{-5}{9}+\dfrac{34}{9}\) 

\(=\dfrac{29}{9}\) 

c) \(6\dfrac{3}{8}-\left(4\dfrac{3}{8}-\dfrac{1}{2}\right)=\dfrac{51}{8}-\dfrac{35}{8}+\dfrac{1}{2}=\left(\dfrac{51}{8}-\dfrac{35}{8}\right)+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\) 

d) \(2\dfrac{1}{3}.1,5-\left(\dfrac{11}{10}+50\%\right):\dfrac{4}{15}\) 

\(=\dfrac{7}{3}.1,5-\dfrac{8}{5}:\dfrac{4}{15}\) 

\(=\dfrac{7}{2}-6\) 

\(=\dfrac{-5}{2}=-2,5\) 

Bài 2:

a) \(\dfrac{3}{2}+x=\dfrac{-5}{3}\) 

           \(x=\dfrac{-5}{3}-\dfrac{3}{2}\) 

           \(x=\dfrac{-19}{6}\) 

b) \(5\dfrac{2}{3}-\left(\dfrac{3}{2}+x\right)=4\dfrac{1}{3}\) 

                   \(\dfrac{3}{2}+x=\dfrac{17}{3}-\dfrac{13}{3}\) 

                   \(\dfrac{3}{2}+x=\dfrac{4}{3}\) 

                          \(x=\dfrac{4}{3}-\dfrac{3}{2}\) 

                          \(x=\dfrac{-1}{6}\) 

b) \(\left|x-\dfrac{1}{2}\right|+75\%=\dfrac{9}{10}\) 

               \(\left|x-\dfrac{1}{2}\right|=\dfrac{9}{10}-75\%\) 

               \(\left|x-\dfrac{1}{2}\right|=\dfrac{3}{20}\) 

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{20}\\x-\dfrac{1}{2}=\dfrac{-3}{20}\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{20}\\x=\dfrac{7}{20}\end{matrix}\right.\)

\(B=\dfrac{10\cdot9}{\sqrt{10}}-2\cdot5+1+2021=9\sqrt{10}-10+2022=9\sqrt{10}+2012\)

=10⋅9/√10−2⋅5+1+2021=9√10−10+2022=9√10+2012

Lời giải:

Nếu $x+y+z+t=0$ thì $M=\frac{-t}{t}=\frac{-x}{x}=\frac{-z}{z}=-1$

$\Rightarrow (M-1)^{2025}=(-1-1)^{2025}=(-2)^{2025}$

Nếu $x+y+z+t\neq 0$. Áp dụng TCDTSBN:

$M=\frac{x+y+z}{t}=\frac{y+z+t}{x}=\frac{z+t+x}{y}=\frac{t+x+y}{z}=\frac{x+y+z+y+z+t+z+t+x+t+x+y}{t+x+y+z}=\frac{3(x+y+z+t)}{x+y+z+t}=3$

$\Rightarrow (M-1)^{2025}=2^{2025}$