1 Kai is tired because he stayed up late watching TV

2 Since I have a broken leg, I fell over while I was playing basketball

3 Sehun is going to be late for school as the bus is late

4 Because Lisa was careless, she broke the cup

5 Rose wants to go home since she feels sick

6 Jimin is hungry as he hasn't eaten all day

7 Since plastic bags are very hard to dissolve, they will cause pollution

8 People reuse and recycle bottles and cans as they want to reduce garbage

9 The sea is becoming increasingly polluted since people drop garbage into the sea

10 We shouldn't throw trash onto the water as polluted water can directly do harm to people's health and kill fish

11 Because of her illness, Lisa didn't go to work yesterday

12 Because of missing the first bus, he came to the office 10 minutes late

13 Because of the bad weather, they canceled the trip to the countryside

14 Because of her richness, she lives happily

15 Because of the heavy rain, we didn't go out

Bạn chia nhỏ nội dung ra để dc trợ giúp nhé

1 D

2 C

3 C

4 A

5 B

6 C

7 D

8 A

9 B

10 C

chữ bé quá

vừa mờ vừa dài :<< bạn dùng tổ hợp phím Window+ Shift + S để chụp một phần của màn hình nhé. Sau đó lưu vào máy r đăng lên

Hướng dẫn:

Ta có:

\(x^2-xy+y^2=2x-3y-2\)

\(\Leftrightarrow2x^2-2xy+2y^2-4x+6y+4+9=9\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+\left(y^2+6y+9\right)=9\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-2\right)^2+\left(y+3\right)^2=9\)

Xét....

Đây là 1 cách nhưng làm hơi dài.

\(x^2-xy+y^2=2x-3y-2\\ \Leftrightarrow x^2-xy+y^2-2x+3y+2=0\left(1\right)\\ \Leftrightarrow x^2-x\left(y+2\right)+y^2+3y+2=0\)

Coi đây là pt bậc 2 ẩn x

Ta có: \(\Delta=\left[-\left(y+2\right)\right]^2-4\left(y^2+3y+2\right)=y^2+4y+4-4y^2-12y-8=-3y^2-8y-4\)

Để pt có nghiệm nguyên thì \(\Delta\ge0\Leftrightarrow-3y^2-8y-4\ge0\Leftrightarrow-2\le y\le-\dfrac{2}{3}\)

\(\Leftrightarrow y\in\left\{-2;-1\right\}\)

Thay y=-2 vào (1) ta có:

\(\left(1\right)\Leftrightarrow x^2-x.\left(-2\right)+\left(-2\right)^2-2x+3.\left(-2\right)+2=0\\ \Leftrightarrow x^2+2x+4-2x-6+2=0\\ \Leftrightarrow x^2=0\Leftrightarrow x=0\)

Thay y=-1 vào pt ta có:

\(\left(1\right)\Leftrightarrow x^2-x.\left(-1\right)+\left(-1\right)^2-2x+3.\left(-1\right)+2=0\\ \Leftrightarrow x^2+x+1-2x-3+2=0\\ \Leftrightarrow x^2-x=0\\ \Leftrightarrow x\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy \(\left(x,y\right)\in\left\{\left(0;-2\right);\left(0;-1\right);\left(1;-1\right)\right\}\)