1 He is told to take a long rest

2 This plant isn't watered everyday

3 Beer is drunk all over the world

4 Uniform is worn to school by students

5 Reports are sent to the manager by the secretary in the afternoon

6 How many languages are spoken in Canada?

Bài 2

1 I was brought this dish by the waiter

2 I wasn't showed the special cameras 

3 Her ticket was showed to her friends

4 The broken cup was hidden in the drawer by Tom

5 The fridge wasmoved into the living room by his uncle

Gọi a là số bị chia, b là số chia ( a thuộc N, a<4000, b>47)

Với b = 48 ta có: a = 82 . 48 + 47 = 3983 < 4000 ( chọn )

Với b = 49 ta có: a = 82 . 49 + 47 = 4065 > 4000 ( loại )

Vậy số chia là 48

1:

#include <bits/stdc++.h>

using namespace std;

long long t,i,n;

int main()

{

cin>>n;

t=0;

for (i=1; i<=n; i++) t+=i;

cout<<t;

return 0;

}

Bài 2: 

#include <bits/stdc++.h>

using namespace std;

long long n,i,t;

int main()

{

cin>>n;

t=0;

for (i=1; i<=n; i++)

if (i%2==0) t+=i;

cout<<t;

return 0;

}

Bài 3: 

#include <bits/stdc++.h>

using namespace std;

long long n,i,t;

int main()

{

cin>>n;

t=0;

for (i=1; i<=n; i++)

if (i%2!=0) t+=i;

cout<<t;

return 0;

}

Bài 4: 

#include <bits/stdc++.h>

using namespace std;

long long n,i,t;

int main()

{

cin>>n;

t=0;

for (i=1; i<=n; i++)

if (i%3==0) t+=i;

cout<<t;

return 0;

}

Bài 5: 

#include <bits/stdc++.h>

using namespace std;

long long n,i,t;

int main()

{

cin>>n;

t=1;

for (i=1; i<=n; i++)

t*=i;

cout<<t;

return 0;

}

\(A=\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+...+\dfrac{10}{2^{10}}\)

\(2A=\dfrac{1}{1}+\dfrac{2}{2}+\dfrac{3}{4}+...+\dfrac{10}{2^9}\)

\(2A-A=\left(1+\dfrac{2}{2}+\dfrac{3}{4}+...+\dfrac{10}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{2}{4}+...+\dfrac{10}{2^{10}}\right)\)

\(A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^9}-\dfrac{10}{2^{10}}\)

\(B=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^9}\)

\(2B=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\)

\(2B-B=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^9}\right)\)

\(B=2-\dfrac{1}{2^9}\)

Suy ra \(A=B-\dfrac{10}{2^{10}}=2-\dfrac{1}{2^9}-\dfrac{10}{2^{10}}=\dfrac{509}{256}\)

bạn tự vẽ hình nha mk vẽ rùi mà lâu quá nên mk xóa lun

a) Cm tam giác DEC đd vs BAC

tam giác DEC và tam giác BAC có

    C : góc chung

   EDC = ABC =90 độ

=> tam giác DEC đd vs tam giác BAC (g-g)

b) Cm AEC đd với BDC

từ câu a => \(\frac{EC}{DC}=\frac{AC}{BC}\) ( 1)

Lại có C : góc chung (2)

Nên từ (1) và (2) => tam giác DEC đồng dạng với tam giác BDC ( c-g)

c) mk nghĩ đã

1C

2A

3C

4B

5C

6D

7B

8A

9D

10C