ride : a horse, a motorbike, a tricycle, an elephent, a bicycle, a donkey, 

drive: a car, a train, a taxi, a truck, a coach, a van, on the highway, home

go: home, abroad, swimming, shopping, sailing, overseas, fishing, jogging

sail: a boat, a yatch, a dighy

travel: abroad, overseas, the ocean, the sea

a: Xét ΔABC vuông tại A và ΔHBA vuông tại H có

góc B chung

=>ΔABC đồng dạng với ΔHBA

=>AB/HB=BC/BC=AC/HA

=>AB*AH=AC*HB

b: AH=căn 5^2-3^2=4cm

BI là phân giác

=>IH/HB=IA/AB

=>IH/3=IA/5=(IH+IA)/(3+5)=4/8=1/2

=>IH=1,5cm; IA=2,5cm

Bài 1:

a: ĐKXĐ: x∉{2;-2}

b: \(A=\left(\frac{1}{2-x}+\frac{3x}{x^2-4}-\frac{2}{x+2}\right):\left(\frac{x^2+4}{4-x^2}+1\right)\)

\(=\left(\frac{-1}{x-2}+\frac{3x}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x+2}\right):\frac{x^2+4+4-x^2}{4-x^2}\)

\(=\frac{-x-2+3x-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\cdot\frac{-\left(x-2\right)\left(x+2\right)}{8}\)

\(=\frac{2x-2-2x+4}{8}\cdot\left(-1\right)=-\frac28=-\frac14\)

=>A không phụ thuộc vào biến

Bài 2:

a: \(f\left(x\right)=x^3-9x^2+27x-27\)

\(=x^3-3x^2-6x^2+18x+9x-27\)

\(=x^2\left(x-3\right)-6x\left(x-3\right)+9\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3\)

b: \(g\left(x\right)=x^2-6x+9\)

\(=x^2-2\cdot x\cdot3+3^2\)

\(=\left(x-3\right)^2\)

=>f(x)⋮g(x)

=>f(x) chia g(x) thì dư 0

\(\frac{f\left(x\right)}{k\left(x\right)}=\frac{x^3-9x^2+27x-27}{x^2-6x+10}\)

\(=\frac{x^3-6x^2+10x-3x^2+18x-30-x+3}{x^2-6x+10}=x-3+\frac{-x+3}{x^2-6x+10}\)

=>f(x) chia k(x) dư -x+3

Áp dụng định lý Ta-lét ta có:
\(\dfrac{AD}{DB}=\dfrac{AE}{EC}\\ \Rightarrow\dfrac{2}{4}=\dfrac{3}{x}\\ \Rightarrow x=3:\dfrac{1}{2}\\ \Rightarrow x=6\left(cm\right)\)

Bạn chụp rõ lại đề ik ạ!

C

1 cooking 

2 listening

3 reading

4 playing

5 cycling

1 People collect a lot of things such as stamps, milk bottle labels, comic books as well as car toys

2 She is 20 years old

3 She is a hairdresser

4 It is collecting candy shells

5 Her friends and relatives

B

1 F

2 F

3 F

4 F

5 T

Câu 1:

a: 5x-2=3x+6

=>5x-3x=2+6

=>2x=8

=>\(x=\dfrac{8}{2}=4\)

b: a<=b

=>-2022a>=-2022b

=>-2022a+2021>=-2022b+2021

Câu 2:

1:

a: ĐKXĐ: x<>1

\(\dfrac{3}{x-1}+1=\dfrac{2x+5}{x-1}\)

=>\(\dfrac{3+x-1}{x-1}=\dfrac{2x+5}{x-1}\)

=>\(2x+5=x+2\)

=>x=-3(nhận)

b: |x-9|=2x-3

=>\(\left\{{}\begin{matrix}2x-3>=0\\\left(2x-3\right)^2=\left(x-9\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\\left(2x-3-x+9\right)\left(2x+3+x-9\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\\left(x+6\right)\left(3x-6\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\\left[{}\begin{matrix}x=-6\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)

=>x=2

2:

\(\dfrac{x-3}{2}-\dfrac{3x+2}{4}< \dfrac{1}{3}\)

=>\(\dfrac{6\left(x-3\right)-3\left(3x+2\right)}{12}< \dfrac{4}{12}\)

=>6x-18-9x-6<4

=>-3x-24<4

=>-3x<28

=>\(x>-\dfrac{28}{3}\)

Câu 3:

Gọi độ dài quãng đường AB là x(km)

(Điều kiện: x>0)

Thời gian đi từ A đến B là \(\dfrac{x}{40}\left(giờ\right)\)

Thời gian ô tô đi từ B về A là \(\dfrac{x}{30}\left(giờ\right)\)

Theo đề, ta có phương trình:

\(\dfrac{x}{40}+\dfrac{x}{30}+\dfrac{1}{2}=9+\dfrac{1}{4}\)

=>\(\dfrac{7x}{120}=8,75\)

=>\(x=8,75:\dfrac{7}{120}=120\cdot1,25=150\left(nhận\right)\)

vậy: Độ dài quãng đường AB là 150km

Bài 5

B=  \(\dfrac{2015}{2016+2017+2018}\)+\(\dfrac{2016}{2016+2017+2018}\)+\(\dfrac{2017}{2016+2017+2018}\)

Ta có:\(\dfrac{2015}{2016}\)>\(\dfrac{2015}{2016+2017+2018}\),\(\dfrac{2016}{2017}\)>\(\dfrac{2016}{2016+2017+2018}\),\(\dfrac{2017}{2018}\)>\(\dfrac{2017}{2016+2017+2018}\)

⇒A>B

Bài 5:

Ta có:

\(B=\dfrac{2015+2016+2017}{2016+2017+2018}\) 

\(B=\dfrac{2015}{2016+2017+2018}+\dfrac{2016}{2016+2017+2018}+\dfrac{2017}{2016+2017+2018}\) 

Vì \(\dfrac{2015}{2016}>\dfrac{2015}{2016+2017+2018}\) 

\(\dfrac{2016}{2017}>\dfrac{2016}{2016+2017+2018}\) 

\(\dfrac{2017}{2018}>\dfrac{2017}{2016+2017+2018}\) 

\(\Rightarrow A>B\)