\(\dfrac{2}{1\cdot6}+\dfrac{2}{11\cdot16}+...+\dfrac{2}{x\left(x+5\right)}=\dfrac{41}{103}\\ =>\dfrac{2}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{x\left(x+5\right)}\right)=\dfrac{41}{103}\\ =>1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{41}{103}:\dfrac{2}{5}=\dfrac{205}{206}\\ =>1-\dfrac{1}{x+5}=\dfrac{205}{206}\\ =>1-\dfrac{1}{x+5}=1-\dfrac{1}{206}\\ =>\dfrac{1}{x+5}=\dfrac{1}{206}\\ =>x+5=206\\ =>x=206-5=201\)

a: ta có: \(\widehat{MAB}=\widehat{ABC}\)

mà hai góc này là hai góc ở vị trí so le trong

nên AM//BC

ta có: \(\widehat{CAN}=\widehat{ACB}\)

mà hai góc này là hai góc ở vị trí so le trong

nên AN//BC

Ta có: AM//BC

NA//BC

mà AM,AN có điểm chung là A

nên M,A,N thẳng hàng

b: Vì M,A,N thẳng hàng nên \(\widehat{MAB}+\widehat{BAC}+\widehat{CAN}=180^0\)

=>\(\widehat{ABC}+\widehat{BAC}+\widehat{ACB}=180^0\)

Sửa đề: \(\frac{49^5+49^7+49^9}{7^{11}+7^{13}+7^{15}+7^{17}+7^{19}+7^{21}}\)

\(=\frac{\left(7^2\right)^5+\left(7^2\right)^7+\left(7^2\right)^9}{7^{11}\left(1+7^2+7^4\right)+7^{17}\left(1+7^2+7^4\right)}\)

\(=\frac{\left(7^{10}+7^{14}+7^{18}\right)}{\left(1+7^2+7^4\right)\left(7^{11}+7^{17}\right)}=\frac{7^{10}\left(1+7^4+7^8\right)}{\left(1+7^2+7^4\right)\cdot7^{11}\left(7^6+1\right)}\)

\(=\frac{1+7^4+7^8}{\left(1+7^2+7^4\right)\cdot7\cdot\left(7^6+1\right)}=\frac{\left(7^8+2\cdot7^4+1\right)-7^4}{\left(7^4+1+7^2\right)\cdot7\cdot\left(7^6+1\right)}\)

\(=\frac{\left(7^4+1\right)^2-7^4}{\left(7^4+7^2+1\right)\cdot7\left(7^6+1\right)}=\frac{\left(7^4+1-7^2\right)\left(7^4+7^2+1\right)}{7\left(7^6+1\right)\left(7^4+7^2+1\right)}\)

\(=\frac{7^4-7^2+1}{7\left(7^6+1\right)}=\frac{7^4-7^2+1}{7\left(7^2+1\right)\left(7^4-7^2+1\right)}=\frac{1}{7\left(7^2+1\right)}=\frac{1}{7\cdot50}=\frac{1}{350}\)

1. more interesting than

2. friendlier/more friendly than

3. more beautiful

4. harder

5. more peaceful - quieter/more quiet

6. more expensive than

7. more convenient than

8. a more exciting

9. faster than

10. harder than

1. Minh says life in the countryside is (interesting) _____more interesting than_____ he expected. He'll go there whenever he has free time.

2. My brother is studying at a university in Ha Noi. He often says he loves living in our village because people here are (friendly) ____friendlier than______ people in Ha Noi.

3. Last week we went to Sa pa. It's (wonderful) ______more wonderful____ to look at the paddy fields on mountain slops than the paddy fields on the lowlands.

4. My grandparents often tell us that they used to live in a (hard) ____harder______ life than it is now.

5. Living in the countryside is (peaceful) __more peaceful________ and (quiet) _____more quiet_____ living in big cities.

\(\left(x+2\right)^2+2\left(y-3\right)^2< 4\)

mà x,y nguyên

nên \(\left[\left(x+2\right)^2;2\left(y-3\right)^2\right]\in\left\{\left(1;2\right);\left(0;2\right)\right\}\)

=>\(\left(x+2;y-3\right)\in\left\{\left(1;1\right);\left(1;-1\right);\left(-1;1\right);\left(-1;-1\right);\left(0;1\right);\left(0;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-1;4\right);\left(-1;2\right);\left(-3;4\right);\left(-3;2\right);\left(-2;4\right);\left(-2;2\right)\right\}\)

a: Ta có: xx'\(\perp\)AB

yy'\(\perp\)AB

Do đó: xx'//yy'

b: xx'//y'y

=>\(\widehat{ADC}=\widehat{C_1}\)(hai góc so le trong)

=>\(\widehat{C_1}=74^0\)

c: DE là phân giác của góc CDF

=>\(\widehat{FDE}=\dfrac{\widehat{FDC}}{2}=\dfrac{106^0}{2}=53^0\)

Xét ΔDEF có \(\widehat{x'FE}\) là góc ngoài tại F

nên \(\widehat{x'FE}=\widehat{FED}+\widehat{FDE}=70^0+53^0=123^0\)

\(\left(x+5\right)^2-4x^2\\=\left(x+5\right)^2-\left(2x\right)^2\\ =\left[\left(x+5\right)-2x\right]\left[\left(x+5\right)+2x\right]\\ =\left(x+5-2x\right)\left(x+5+2x\right)\\ =\left(-x+5\right)\left(3x+5\right)\)

`a` là số tự nhiên không chia hết cho `3` nên a có dạng: 

`a = 3k + 1` hoặc `a = 3k + 2`

(`k` thuộc `N`*)

Mà a là số tự nhiên lẻ `=> a^2` là số tự nhiên lẻ `=> a^2 - 1` là số chẵn 

`=> a^2 ⋮ 2`

Để `a^2 - 1 ⋮ 6` thì  `a^2 - 1 ⋮ 3` (Vì `UCLN(2;3) = 1`)

- Xét `a = 3k + 1`

`=> a^2 -1 = (3k+1)^2 -1= 9k^2 + 6k + 1 - 1= 9k^2 + 6k^2 ⋮ 3` (Thỏa mãn)

- Xét `a = 3k + 2`

`=> a^2 -1 = (3k+2)^2 -1 = 9k^2 + 12k + 4 - 1= 9k^2 + 12k^2 + 3 ⋮ 3` (Thỏa mãn)

Vậy ...