\(\dfrac{x^3+8}{x^2+2x+1}.\dfrac{x^2+3x+2}{1-x^2}\left(x\ne\pm1\right)\\ =\dfrac{x^3+2^3}{\left(x+1\right)^2}.\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{1^2-x^2}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{\left(x+2\right)\left(x+1\right)}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(1-x\right)\left(x+1\right)^2}\)

x(x + 2) + x(x - 5) - 5(x + 2)

= [x(x + 2) - 5(x + 2)] + x(x - 5)

= (x - 5)(x + 2) + x(x - 5)

= (x - 5)(x + 2 + x)

= (x - 5)(2x + 2)

= 2(x - 5)(x + 1)

a,A =  \(\dfrac{3x^2+6xy}{6x^2}\)  (đk  \(x\) ≠ 0)

  A = \(\dfrac{3x.\left(x+2y\right)}{6x^2}\)

 A = \(\dfrac{x+2y}{2x}\) 

b,B =  \(\dfrac{2x^2-x^3}{x^2-4}\) (đk \(x\)²  - 4 ≠ 0 ⇒ \(x\) ≠ \(\pm\) 2)

   B =  \(\dfrac{x^2\left(2-x\right)}{\left(x+2\right)\left(x-2\right)}\)

   B = \(\dfrac{-x^2.\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}\)

   B =  \(\dfrac{-x^2}{x+2}\)

a) \(\left(x^2-4x\right)+\left(12x-48\right)\)

\(=x\left(x-4\right)+12\left(x-4\right)\)

\(=\left(x-4\right)\left(x+12\right)\)

b) \(x(x+y)-3x-3y\)

\(=x\left(x+y\right)-\left(3x+3y\right)\)

\(=x\left(x+y\right)-3\left(x+y\right)\)

\(=\left(x+y\right)\left(x-3\right)\)

c) \(20xy^2-12x^2+12x^3y\)

\(=4x\left(5y^2-3x+3x^2y\right)\)

d) \(2x\left(x-y\right)+3y\left(y-x\right)\)

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

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

\(9(x-y)^2-27(y-x)^3\\=9(x-y)^2+27(x-y)^3\\=(x-y)^2[9+27(x-y)]\\=(x-y)^2(9+27x-27y)\\=9(x-y)^2(1+3x-3y)\)

a) \(70a+84b-20ab-24b^2\)

\(=\left(70a+84b\right)-\left(20ab+24b^2\right)\)

\(=14\left(5a+6b\right)-4b\left(5a+6b\right)\)

\(=\left(5a+6b\right)\left(14-4b\right)\)

\(=2\left(5a+6b\right)\left(7-2b\right)\)

b) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)

\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xyz+xz^2\right)+\left(xyz+y^2z+yz^2\right)\)

\(=xy\left(x+y+z\right)+xz\left(x+y+z\right)+yz\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(xy+yz+xz\right)\)

c) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)

\(=\left(x^2y+xy^2\right)+\left(xz^2+yz^2\right)+\left(x^2z+2xyz+y^2z\right)\)

\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x^2+2xy+y^2\right)\)

\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x+y\right)^2\)

\(=\left(x+y\right)\left[xy+z^2+z\left(x+y\right)\right]\)

\(=\left(x+y\right)\left(xy+z^2+xz+yz\right)\)

\(=\left(x+y\right)\left[\left(xy+yz\right)+\left(xz+z^2\right)\right]\)

\(=\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]\)

\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

a, 70a + 84b - 20ab - 24b²

 = 14.(5a + 6b) - 4b(5a + 6b)

= (5a + 6b).(14 - 4b) 

a)Áp suất ở dưới pittong nhỏ là: \(\dfrac{10m_2}{S_2}=\dfrac{10m_1}{S_1}+10D\cdot h\)

\(\Rightarrow\dfrac{10m_2}{25\cdot10^{-4}}=\dfrac{10\cdot1}{50\cdot10^{-4}}+10\cdot1000\cdot0,1\Rightarrow m_2=0,75kg=750g\)

b)Khi đặt lên pittong bên trái một lượng \(m=300g=0,3kg\) thì nó di chuyển xuống dưới một đoạn:

\(\dfrac{10\left(m_2+m\right)}{S_2}=\dfrac{10m_1}{S_1}+10D\cdot\Delta h\)

\(\Rightarrow\dfrac{10\cdot\left(0,75+0,3\right)}{25\cdot10^{-4}}=\dfrac{10\cdot1}{50\cdot10^{-4}}+10\cdot1000\cdot\Delta h\)

\(\Rightarrow\Delta h=0,22m=22cm\)

1 The man is a monk from Emei Mountain

2 They make sacrificial offerings to the gods

3 The ornamental tree in our front yard is growing rapidly

4 Martial arts originates in the East

5 Young rice cake is a specialty of this area.

6 You should learn Vietnamese table manners

7 We have a family reunion next week