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

Câu 6: Câu nói của Nhím đối với Thỏ "Thế thì gay go đấy! Trời rét, không có áo khoác thì chịu sao được" thể hiện thái độ cảm thông, xót xa của Nhím trước tình cảnh thiếu thốn của Thỏ. Qua đó chúng ta có thể thấy Nhím là người giàu lòng nhân ái, yêu thương với những hoàn cảnh khó khăn

Chọn B

28.A

29.C

30.B

31.B

16, Jane hasn't cooked for her family since she was at high school

17, The last time Jane cooked for her family was when she was at high school

1B

2C

3C

4C

5D

6C

7D

Ta có: \(-3x^2-5x-2=0\)

Theo định lý vi-et ta có: 

\(x_1+x_2=-\dfrac{b}{a}=-\dfrac{-5}{-3}=-\dfrac{5}{3}\)

\(x_1x_2=\dfrac{c}{a}=\dfrac{-2}{-3}=\dfrac{2}{3}\) 

a) \(M=x_1+\dfrac{1}{x_1}+\dfrac{1}{x_2}+x_2\)

\(M=\left(x_1+x_2\right)+\dfrac{x_1+x_2}{x_1x_2}\)

\(M=-\dfrac{5}{3}+\dfrac{-\dfrac{5}{3}}{\dfrac{2}{3}}=-\dfrac{25}{6}\)

b) \(N=\dfrac{1}{x_1+3}+\dfrac{1}{x_2+3}\)

\(N=\dfrac{x_2+3+x_1+3}{\left(x_1+3\right)\left(x_2+3\right)}\)

\(N=\dfrac{\left(x_1+x_2\right)+6}{x_1x_2+3\left(x_1+x_2\right)+9}\)

\(N=\dfrac{-\dfrac{5}{3}+6}{\dfrac{2}{3}+3\cdot-\dfrac{5}{3}+9}=\dfrac{13}{14}\) 

c) \(P=\dfrac{x_1-3}{x^2_1}+\dfrac{x_2-3}{x^2_2}\)

\(P=\dfrac{x^2_2\left(x_1-3\right)+x^2_1\left(x_2-3\right)}{x^2_1x^2_2}\)

\(P=\dfrac{x^2_2x_1+x^2_1x_2-3x^2_2-3x^2_1}{\left(x_1x_2\right)^2}\)

\(P=\dfrac{x_1x_2\left(x_1+x_2\right)-3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}{\left(x_1x_2\right)^2}\)

\(P=\dfrac{\dfrac{2}{3}\cdot-\dfrac{5}{3}-3\cdot\left[\left(-\dfrac{5}{3}\right)^2-2\cdot\dfrac{2}{3}\right]}{\left(\dfrac{2}{3}\right)^2}=-\dfrac{49}{4}\) 

d) \(Q=\dfrac{x_1}{x_2+2}+\dfrac{x_2}{x_1+2}\)

\(Q=\dfrac{x_1\left(x_1+2\right)+x_2\left(x_2+2\right)}{\left(x_2+2\right)\left(x_1+2\right)}\)

\(Q=\dfrac{x^2_1+2x_1+x_2^2+2x_2}{x_1x_2+2x_2+2x_1+4}\)

\(Q=\dfrac{\left(x^2_1+x^2_2\right)+2\left(x_1+x_2\right)}{x_1x_2+2\left(x_1+x_2\right)+4}\)

\(Q=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2+2\left(x_1+x_2\right)}{x_1x_2+2\left(x_1+x_2\right)+4}\)

\(Q=\dfrac{\left(-\dfrac{5}{3}\right)^2-2\cdot\dfrac{2}{3}+2\cdot-\dfrac{5}{3}}{\dfrac{2}{3}+2\cdot-\dfrac{5}{3}+4}=-\dfrac{17}{12}\)

Bài 2:

\(x^2+\left(m+2\right)x+2m=0\)

\(\text{Δ}=\left(m+2\right)^2-4\cdot1\cdot2m\)

\(=m^2+4m+4-8m=m^2-4m+4\)

\(=\left(m-2\right)^2>=0\forall m\)

=>Phương trình luôn có hai nghiệm x1;x2

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(m+2\right)}{1}=-m-2\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{2m}{1}=2m\end{matrix}\right.\)

\(2\cdot\left(x_1+x_2\right)+x_1x_2\)

\(=2\left(-m-2\right)+2m\)

=-2m-4+2m

=-4

=>Đây là hệ thức cần tìm

Bài 3:

a: Thay x=-2 vào phương trình, ta được:

\(\left(2m-1\right)\cdot\left(-2\right)^2+\left(m-3\right)\cdot\left(-2\right)-6m-2=0\)

=>\(4\left(2m-1\right)-2\left(m-3\right)-6m-2=0\)

=>8m-4-2m+6-6m-2=0

=>0=0

=>Phương trình luôn có nghiệm x=-2

b: TH1: m=1/2

Phương trình lúc này sẽ là:

\(\left(2\cdot\dfrac{1}{2}-1\right)\cdot x^2+\left(\dfrac{1}{2}-3\right)x-6\cdot\dfrac{1}{2}-2=0\)

\(\Leftrightarrow-\dfrac{5}{2}x-5=0\)

=>\(-\dfrac{5}{2}x=5\)

=>\(x=-5:\dfrac{5}{2}=-2\)

TH2: m<>1/2

\(\text{Δ}=\left(m-3\right)^2-4\left(2m-1\right)\left(-6m-2\right)\)

\(=m^2-6m+9+4\left(2m-1\right)\left(6m+2\right)\)

\(=m^2-6m+9+4\left(12m^2+4m-6m-2\right)\)

\(=m^2-6m+9+4\left(12m^2-2m-2\right)\)

\(=m^2-6m+9+48m^2-8m-8\)

\(=49m^2-14m+1=\left(7m-1\right)^2>=0\forall m\)

=>Phương trình luôn có hai nghiệm là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-\left(m-3\right)-\sqrt{\left(7m-1\right)^2}}{2\cdot\left(2m-1\right)}=\dfrac{-\left(m-3\right)-\left|7m-1\right|}{4m-2}\\x_2=\dfrac{-\left(m-3\right)+\sqrt{\left(7m-1\right)^2}}{2\left(2m-1\right)}=\dfrac{-\left(m-3\right)+\left|7m-1\right|}{4m-2}\end{matrix}\right.\)