Trong 14 số tự nhiên có 3 chữ số chắc chắn có 2 số chia cho 13 có cùng số dư 
Nên hiệu của chúng chia hết cho 13 
Gọi số có 6 chữ số chia hết cho 13 là abcdeg (có gạch trên đầu) thì abc-deg chia hết cho 13 
Ta có: abcdeg + (abc-deg) 
= abcdeg + abc-deg 
= 1000.abc + deg + abc - deg 
= (1000+1).abc + (deg-deg) 
= 1001.abc + 0 
= 1001.abc 
Vì 1001 chia hết cho 13 nên 1001.abc cũng chia hết cho 13 
=> abcdeg + (abc-deg) chia hết cho 13 
Mà abc-deg chia hết cho 13 
Nên abcdeg chia hết cho 13 
Vây trong 14 số đó tồn tại 2 số mà khi viết liên nhau thì tạo thành số có 6 chữ số chia hết cho 13

Bài 1: Choose the correct word to complete the sentences

1. I haven't you spoken to her.......(for/since/ago) she went to London

2. He left school three years.......(for/since/ago), then he worked in France......(for/since/ago) a few months before coming back to England

3. She has got a headache.........(for/since/ago) she got up this morning

4. I have been here....(for/since/ago) an hour. Where have you been?

A = -1² + 2² - 3² + 4² - ... - 99² + 100²

= (2²-1²) + (4²-3²) + ...+ (100²-99²)

= (2-1)(1+2) + (4-3)(3+4) + ... + (100-99)(99+100)

= 1 + 2 + 3 + ... + 100

= \(\dfrac{100.101}{2}=5050\)

Vậy A = 5050

A = |2x-1| + 5 \(\ge\)5

=> MinA = 5

Dấu "=" xảy ra khi x = \(\dfrac{1}{2}\)

Chia các thì ở dạng đúng trong các câu sau :

1.She (go) goesto school everyday

2.There (to be) are to big problems in many part of the

3.A lost of people (go) go to work by car

4.Davis never (take) takes the bus to school .

5.He plays volleyball but he (not enjoy) doesn't ejnoy it

6 Most shops usually (open) open at 8.00 a.m and (close) close at 6.00 p.m

7.School children (wear) wear uniform on Monday

8. I want (buy) to buy some stamps

9.Ruth want (live) to livewith his parnets in an apartment

a) (5x³-x+2)(x-1) = 5x⁴-5x³-x²+3x-2

b) (4x+4)(3-x²-x³) = 12x-4x³-4x⁴+12-4x²-4x³ = -4x⁴ -8x³ - 4x² + 12x +12

\(G=4a^2+b^2-\text{4ab+4a-2b}\)

\(\Rightarrow G=\left(2a-b\right)^2+2\left(2a-b\right)\)=\(\left(2a-b\right)\left(2a-b+2\right)\)

\(\Rightarrow G=\left(a+a-b\right)\left(a+a-b+2\right)\)

Thay a-b=1 vào \(\Rightarrow G=\left(a+1\right)\left(a+3\right)\Rightarrow G=\left(a+2-1\right)\left(a+2+1\right)\)

\(\Rightarrow G=\left(a+2\right)^2-1\ge-1\Rightarrow MinG=-1\)

Dấu "=" xảy ra khi x=-2

Vậy Min G=-1

a) (x-2)³ = (x-2)²

<=> (x-2)³-(x-2)² = 0

<=> (x-2)²(x-2-1) = 0

<=> \(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\x-3=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

b) \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+...+\dfrac{3x}{11.14}=\dfrac{1}{21}\)

<=> \(x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)

<=> \(x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

<=> \(x=\dfrac{1}{21}:\dfrac{3}{7}\)

<=> \(x=\dfrac{1}{9}\)

Tác dụng của áp lực càng lớn khi áp lực càng mạnh và diện tích bị ép càng hẹp

a.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (1)

\(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\dfrac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\)(2)

Từ (1) và (2) suy ra: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

b.M = \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{50^2}\right)\)

= \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{2499}{2500}\)

= \(\dfrac{1.3.2.4.3.5...49.51}{2^2.3^2.4^2...50^2}\)

\(\dfrac{51}{2.50}=\dfrac{51}{100}\)