Bài 9:
Để A là số nguyên thì \(4x-10⋮x-2\)

=>\(4x-8-2⋮x-2\)

=>\(-2⋮x-2\)

=>\(x-2\inƯ\left(-2\right)\)

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

=>\(x\in\left\{3;1;4;0\right\}\)

Bài 8:

Diện tích mảnh vườn là:

\(\dfrac{1}{2}\cdot10\cdot\left(15+25\right)=5\cdot40=200\left(m^2\right)\)

Khối lượng thóc thu được là:

\(200:1\cdot0,7=140\left(kg\right)\)

Bài 7:

\(\dfrac{2}{3}\cdot\left(\dfrac{1}{15}-\dfrac{5}{7}\right)+\dfrac{2}{3}+\dfrac{2}{3}\cdot\left(\dfrac{14}{15}+\dfrac{19}{7}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{1}{15}-\dfrac{5}{7}+1+\dfrac{14}{15}+\dfrac{19}{7}\right)\)

\(=\dfrac{2}{3}\left(1+1+2\right)\)

\(=\dfrac{2}{3}\cdot4=\dfrac{8}{3}\)

Bài 6:

\(\left(-x+0,2\right)^3=0,008\)

=>\(-x+0,2=\sqrt[3]{0,008}=0,2\)

=>-x=0

=>x=0

=>Có 1 giá trị x thỏa mãn

Câu 4:

\(\left(\dfrac{2x}{3}-3\right):\left(-10\right)=\dfrac{2}{5}\)

=>\(x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)

=>\(x\cdot\dfrac{2}{3}=-4+3=-1\)

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

23.17-23.14

=23.(17-14)

=23.3 = 69

23.(17-14)=23.3=69

\(A=x^6-2x^4-2x^4+4x^2+2x^3-4x\\ A=x^3\left(x^3-2x\right)-2x\left(x^3-2x\right)+2\left(x^3-2x\right)\\ A=\left(x^3-2x\right)\left(x^3-2x+2\right)=3\left(3+2\right)=3\cdot5=15\\ B=x^5-2x^3+3x^3-6x-3x^2\\ =x^2\left(x^3-2x\right)+3\left(x^3-2x\right)-3x^2\\ =\left(x^3-2x\right)\left(x^2+3\right)-3x^2=3\left(x^2+3\right)-3x^2\\ =3x^2-3x^2+9=9\)

Bài 6:

a: Xét ΔABM và ΔACM có 

AB=AC

AM chung

BM=CM

Do đó: ΔABM=ΔACM

Ta có: ΔABC cân tại A

mà AM là đường trung tuyến

nên AM là đường phân giác

b: Xét ΔADM và ΔAEM có 

AD=AE

\(\widehat{DAM}=\widehat{EAM}\)

AM chung

Do đó: ΔADM=ΔAEM

Suy ra: \(\widehat{ADM}=\widehat{AEM}=90^0\)

hay ME⊥AC

Bài 2

1 a lot of

2 much

3 a lot of - many

4 many

5 much

Bài 3

1 much

2 much

3 a lot of

4 lots of

5 many

Bài 4

1 much

2 many

3 a lot of

4 many

5 lots of

Bài 5

1 any - a - many

3 any

4 lots

5 many - lot

6 lot - any

III

1 desks

2 students

3 televisions

4 couches

5 bookshelves