Nhanh giúp mk với ạ

A

1. Line 1: "In England, when the school children come to school, they first go to the cloakroom."  
   - Redundant word: "school" (appears twice in the line)

2. Line 2: "They are take off their coats and raincoats, their caps and hats, and then go to their classroom."  
   - Redundant word: "are" (the correct phrase should be "They take off their coats...")

3. Line 3: "Some of the students go for to the laboratories and workshop where they learn in physics, chemistry, and art."  
   - Redundant word: "for" (the correct phrase should be "go to the laboratories...")

4. Line 4: "When a students is on duty, he comes to school very early."  
   - Redundant word: "a" (should be "When a student is on duty...")

5. Line 5: "He or she has to open all the window, water in the flower and clean the blackboard, so all everything is ready."  
   - Redundant word: "all" (should be "so everything is ready.")

line 1: school (1)

line 2: are 

line 3: their

line 4: for

line 5: in

line 6: students => student

line 7: very 

line 8: in

line 9: all

line 10: to

Here are the answers based on the passage:

1. Where is Michael's school?

   • Michael's school is in California.

2. How long does it take him to go to school?

   • It takes Michael about 15 minutes to go to school.

3. What time do his classes end?

   • His classes end at 11:15.

4. What does he do in the afternoon?

   • In the afternoon, Michael plays sports and learns music.

5. What time does he go to bed?

   • He and his family often go to bed at 10 o'clock.

1 It is in California

2 It takes him 15 minutes to go to school

3 They end at 11:15

4 He plays sports and learns music

5 He goes to bed at 10

.............

Bạn cần giúp j nhỉ

\(\dfrac{4}{x}+\dfrac{2}{y}=1\) \(\left(x;y\ne0\right)\)

\(\Rightarrow2x+4y=xy\)

\(\Rightarrow2x-8+4y-xy=-8\)

\(\Rightarrow2\left(x-4\right)-y\left(x-4\right)=-8\)

\(\Rightarrow\left(x-4\right)\left(y-2\right)=8\)

\(\Rightarrow\left(x-4;y-2\right)\in U\left(8\right)=\left\{-1;1;-2;2;-4;4;-8;8\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(3;-6\right);\left(5;10\right);\left(2;-2\right);\left(8;4\right);\left(-4;;1\right);\left(12;3\right)\right\}\)

Ta có : a chia 6 dư 2 => a - 2 chia hết cho 6 => a - 2 + 12 chia hết cho 6 => a + 10 chia hết cho 6

a chia 7 dư 4 => a - 4 chia hết cho 7 => a - 4 + 14 chia hết cho 7 => a + 10 chia hết cho 7

=> a + 10 chia hết cho 6 và 7

=. a + 10 thuộc BC ( 6 ; 7 )

Mà BCNN ( 6 ; 7 ) = 42

=> a + 10 thuộc B ( 42 ) = { 0 ; 42 ; ... }

=> a + 10 chia 42 dư 42

=> a chia 42 dư 32

Vậy số a chia cho 42 dư 32

Với k = 1 ta có:

A =  6k + 5 = 6.1 + 5 = 11 (là số nguyên tố)

Vậy tồn tại số nguyên tố dạng: 6k + 5 (đpcm)