2ax² - bx² - 2ax + bx + 4a - 2b

= (2ax² - 2ax + 4a) - (bx² - bx + 2b)

= 2a.(x² - x + 2) - b.(x² - x + 2)

= (x² - x + 2).(2a - b)

= (x² - 2x + x + 2).(2a - b)

= [-x.(x + 2) + (x + 2)].(2a - b)

= (x + 2).(1 - x).(2a - b)

1) (a - b)x - (b - a)y + a - b

= (a - b)x + (a - b)y + (a - b)

= (a - b)(x + y + 1)

2) (x - y + z)a + (y - x - z)b - x + y - z

= (x - y + z)a - (x - y + z)b - (x - y + z)

= (x - y + z)(a - b - 1)

3) (2a + 3)x - 2ay - 3y + 2a + 3

= (2a + 3)x - (2a + 3)y + (2a + 3)

= (2a + 3)(x - y + 1)

4) (a - b)x + by - ay - a +b

= (a - b)x - (a - b)y - (a - b)

= (a - b)(x - y - 1)

c) (ab + 1)² - (a + b)²

= (ab + 1 - a - b).(ab + 1 + a + b)

d) x² - 2x - 4y² - 4y

= (x² - (2y)²) - (2x + 4y)

= (x - 2y).(x + 2y) - 2.(x + 2y)

= (x + 2y).(x - 2y - 2)

e) x² - y² + 2x + 1

= (x² + 2x + 1) - y²

= (x + 1)² - y²

= (x + y + 1).(x - y + 1)

f) x³ + 2x² + 2x + 1

= x³ + 2.x².1 + 2.x.1² + 1³

= (x + 1)³

g) 8x³ - 27y³

= (2x)³ - (3y)³

= (2x - 3y).(4x² + 6xy + 9y²)

h) x³ + y³ + 2x² - 2xy + 2y²

= (x³ + y³) + (2x² - 2xy + 2y²)

= (x + y).(x² - xy + y²) + 2.(x² - xy + y²)

= (x² - xy + y²).(x + y + 2)

i) x² + y² - 2xy - 4x + 4y

= (x² - 2xy + y²) - (4x - 4y)

= (x - y)² - 4.(x - y)

= (x - y).(x - y - 4)

a) \(6x^2+11x+3\)

\(=6x^2+2x+9x+3\)

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

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

b) x² + 6x + 5

= x² + x + 5x + 5

= x.(x + 1) + 5.(x + 1)

= (x + 1).(x + 5)

c) x² - 7x + 10

= x² - 2x - 5x + 10

= x.(x - 2) - 5.(x - 2)

= (x - 2).(x - 5)

a) 2x² - 28x + 98

= 2.(x² - 2.x.7 + 7²)

= 2.(x - 7)²

b) 5x² - 20y²

= 5.[x² - (2y)² ]

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

c) 4x² - 324

= 4.(x² - 9²)

= 4.(x - 9).(x + 9)

d) 3a² + 6ab + 3b²

= 3.(a² + 2ab + b²)

= 3.(a + b)²

e) (a - 4)² - 4y²

= (a - 4)² - (2y)²

= (a - 2y - 4).(a + 2y - 4)

f) 2x³ + 16

= 2.(x³ + 2³)

= 2.(x + 2).(x² + 2x + 4)

\(n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02\left(mol\right)\)

500 ml = 0,5 l

\(n_{H_2SO_4}=0,2.0,5=0,1\left(mol\right)\)

a) PT: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Trước 0,02 0,1 0 0 mol

Trong 0,02 0,06 0,02 0,06 mol

Sau 0 0,04 0,02 0,06 mol

b) \(m_{Fe_2\left(SO_4\right)_3}=0,02.400=8\left(g\right)\)

c) \(C_{MH_2SO_4dư}=\dfrac{0,04}{0,5}=0,08M\)

\(m_{BaCl_2}=\dfrac{200.10,4}{100}=20,8\left(g\right)\)

\(n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

\(a.BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

mol 0,1 → 0,1 0,1 0,2

b.\(m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

\(m_{HCl}=0,2.36,5=7,3\left(g\right)\)