\(\left(-\dfrac{3}{5}\right)^2.\dfrac{5}{11}+\dfrac{9}{25}.\left(-\dfrac{16}{11}\right)\)

\(=\dfrac{9}{25}.\dfrac{5}{11}+\dfrac{9}{25}.\left(-\dfrac{16}{11}\right)\)

\(=\dfrac{9}{25}.\left[\dfrac{5}{11}+\left(-\dfrac{16}{11}\right)\right]\)

\(=\dfrac{9}{25}.\left(-1\right)\)

\(=-\dfrac{9}{25}\)

Ta có: \(\widehat{xOt}+\widehat{xOy}=180^0\)(hai góc kề bù)

=>\(\widehat{xOt}=180^0-70^0=110^0\)

Ta có: \(\widehat{xOt}=\widehat{yOz}\)(hai góc đối đỉnh)

mà \(\widehat{xOt}=110^0\)

nên \(\widehat{yOz}=110^0\)

Ta có: \(\widehat{yOz}+\widehat{zOt}=180^0\)(hai góc kề bù)

=>\(\widehat{zOt}=180^0-110^0=70^0\)

Do `Oz` đối tia `Ox`

=> \(\widehat{xOz}=180^o\)

Mà \(\widehat{xOy}+\widehat{yOz}=\widehat{xOz}\) (Vì `Oy` nằm giữa `Oz` và `Ox`)

=> \(70^o+\widehat{yOz}=180^o\)

=> \(\widehat{yOz}=180^o-70^o\)

=> \(\widehat{yOz}=110^o\)

Lại có: \(\left\{{}\begin{matrix}\widehat{xOy}=\widehat{tOz}\\\widehat{yOz}=\widehat{xOt}\end{matrix}\right.\) (Các cặp góc đối đỉnh)

=> \(\left\{{}\begin{matrix}\widehat{zOt}=70^o\\\widehat{xOt}=110^o\end{matrix}\right.\)

Vậy ...

7\(x\).2 hay 7\(x^2\) vậy em?

7x^2 ạ

Ta có:

∠A₃ = ∠B₁ (gt)

Mà ∠A₃ và ∠B₁ là hai góc so le trong

⇒ a // b

⇒ ∠A₂ = ∠B₂ (đồng vị)

Do ∠A₃ + ∠A₄ = 180⁰ (kề bù)

Mà ∠A₃ = ∠B₁ (gt)

⇒ ∠B₁ + ∠A₄ = 180⁰

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

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

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

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