1. Someone checks the water level every week

The water level is checked every week.

2. We expect students not to talk during ypthe examination

Students are expected not to talk during ypthe examination.

3. We invited two hundred people to the party

Two hundred people were invited to the party.

4. He is mending your shoes at the moment

Your shoes are being mended by him at the moment.

5. We don't allow smoking in this restaurant

Smoking is not allowed in this restaurant.

6. Someone is interviewing Dr.Smiths at the moment

Dr. Smiths is being interviewed at the moment.

7. You must clean this machine every time it you use it.

This machine must be cleaned every time it is used.

8. We had warned him not to go near the river the day before

He had been warned not to go near the river the day before.

9. Someone has moved my chair

My chair has been moved.

10. They were repairing the machine when the accident happened

The machine was being repaired when the accident happened.

Xét tam giác ABC và tam giác BAD có: 

AC=BD( gt)

BC=AD( gt)

AB chung

=> △ABC=△BAD( c.c.c)

=> \(\widehat{ABC}=\widehat{BAD}\) (2 góc tương ứng)

Mà 2 góc này là 2 góc so le trong

=> AC//BD

a) \(a^2\left(a+1\right)+2a\left(a+1\right)=\left(a+1\right)\left(a^2+2a\right)=a\left(a+1\right)\left(a+2\right)\) là tích 3 số nguyên liên tiếp nên có một số chia hết cho 2 và một số chia hết cho 3

\(\Rightarrow a^2\left(a+1\right)+2a\left(a+1\right)⋮2.3=6,\forall a\in Z\)

b) \(a\left(2a-3\right)-2a\left(a+1\right)=2a^2-3a-2a^2-2a=-5a⋮5,\forall a\in Z\)

c) \(\left(2a-1\right)^3-\left(2a-1\right)=\left(2a-1\right)\left[\left(2a-1\right)^2-1\right]=\left(2a-1\right)\left(2a-1-1\right)\left(2a-1+1\right)=\left(2a-2\right)\left(2a-1\right)2a=4\left(2a-1\right)\left(a-1\right)a\)

Do \(\left(a-1\right)a\) là tích 2 số nguyên liên tiếp nên chia hết cho 2 \(\Rightarrow4a\left(a-1\right)⋮8\Rightarrow4\left(2a-1\right)\left(a-1\right)a⋮8\Rightarrow\left(2a-1\right)^3-\left(2a-1\right)⋮8,\forall a\in Z\)

d) \(a^3-19a=a^3-a-18a=a\left(a^2-1\right)-18a=\left(a-1\right)a\left(a+1\right)-18a\)

Ta có \(\left(a-1\right)a\left(a+1\right)⋮6,-18a⋮6,\forall a\in Z\)

Nên suy ra \(a^3-19a⋮6,\forall a\in Z\)

a) Gọi 3 số chẵn liên tiếp lần lượt là: \(2x,2x+2,2x+4\)

Theo đề bài ta có phương trình:

       \(\left(2x+2\right)\left(2x+4\right)-2x\left(2x+2\right)=192\Rightarrow4x^2+12x+8-4x^2-4x=192\Rightarrow8x=184\Rightarrow x=23\)

Vậy 3 số chẵn liên tiếp đó lần lượt là: 46, 48, 50

b) Gọi 4 số tự nhiên liên tiếp đó lần lượt là: x, x+1, x+2, x+3

Theo đề bài ta có phương trình: 

\(\left(x+2\right)\left(x+3\right)-x\left(x+1\right)=146\Rightarrow x^2+5x+6-x^2-x=146\Rightarrow4x=140\Rightarrow x=35\)

Vậy 4 số tự nhiên liên tiếp đó lần lượt là: 35,36,37,38

a) \(a\left(a-1\right)-\left(a+3\right)\left(a+2\right)=a^2-a-a^2-5a-6=-6a-6=-6\left(a+1\right)⋮6,\forall a\in Z\)

b) \(a\left(a+2\right)-\left(a-5\right)\left(a-7\right)=a^2+2a-a^2+12a-35=14a-35=7\left(2a-5\right)⋮7,\forall a\in Z\)c) \(\left(n^2-3n+1\right)\left(n+2\right)-n^3+n^2+3=n^3-n^2-5n+2-n^3+n^2+3=-5n+5=-5\left(n-5\right)⋮5,\forall n\in Z\)

\(\sqrt{2x^2+4x+5}=\sqrt{2\left(x^2+2x+1\right)+3}=\sqrt{2\left(x+1\right)^2+3}\)

Do \(2\left(x+1\right)^2+3\ge3>0\forall x\) nên điều kiện xác định của x là: \(x\in R\)

Số nghịch đảo của -0,125 là \(\dfrac{1}{-0,125}=-8\)

1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)

\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)

Nên suy ra \(10A>10B\Rightarrow A>B\)

2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

              \(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)

Vậy \(2^{332}< 3^{223}\)