D

Ta có : (x+2)(x+4)(x+6)(x+8) + 16

=[(x+2).(x+8)].[(x+4)(x+6)]+16

=(x²+10x+16).(x²+10x+24)+16 (1)

Đặt x^2+10x+16=a thì (1) trở thành:

a.(a+8)+16=a²+8a+16=(a+4)²=(x^2+10x+20)²

1.A

2.C

3.B

4.C

a

c

b

c

a) x² - 4x + 3
= x² - x - 3x + 3
= x(x - 1) - 3(x - 1)
= (x - 1)(x - 3)
b) x² + 5x + 4
= x² + x + 4x + 4
= x(x + 1) + 4(x + 1)
= (x + 1)(x + 4)
c) x² - x - 6
= x² + 2x - 3x - 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)

x² - 4x + 3

= x² - x - 3x + 3

= (x² - x) - (3x - 3)

= x(x - 1) - 3(x - 1)

= (x - 3) (x - 1)

vại

fdvfdverberrgtrgrgg

\(x^4+2018x^2+2017x+2018\)

\(=\left(x^4-x\right)+\left(2018x^2+2018x+2018\right)\)

\(=x.\left(x^3-1\right)+2018.\left(x^2+x+1\right)\)

\(=x.\left(x-1\right)\left(x^2+x+1\right)+2018.\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2018\right)\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{a}{3}=\frac{b}{4}=\frac{a+b}{7}=\frac{c}{5}=\frac{d}{6}=\frac{c-d}{-1}\)

\(\frac{a+b}{7}=\frac{c-d}{-1}\Rightarrow\frac{a+b}{c-d}=-7\)

Bài 1 yêu cầu gì em?

Bài 2:

\(a,x\left(x-1\right)+5\left(x-1\right)=\left(x+5\right)\left(x-1\right)\\ b,3x\left(x+1\right)+3\left(x+1\right)=\left(3x+3\right)\left(x+1\right)=3\left(x+1\right)\left(x+1\right)=3\left(x+1\right)^2\\ c,x\left(x-3\right)+xy\left(x-3\right)=\left(x+xy\right)\left(x-3\right)=x\left(y+1\right)\left(x-3\right)\\ d,2x\left(x-2\right)-6\left(x-2\right)=\left(2x-6\right)\left(x-2\right)=2\left(x-3\right)\left(x-2\right)\)

Bài 1:

a) \(3xy+6y\)

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

b) \(3x^2+9x\)

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

c) \(6x-9y^2\)

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

d) \(10xy^2-6x^2y\)

\(=2xy\left(5y-3x\right)\)

Bài 2:

a) \(x\left(x-1\right)+5\left(x-1\right)\)

\(=\left(x-1\right)\left(x+5\right)\)

b) \(3x\left(x+1\right)+3\left(x+1\right)\)

\(=\left(x+1\right)\left(3x+3\right)\)

\(=3\left(x+1\right)\left(x+1\right)\)

\(=3\left(x+1\right)^2\) 

c) \(x\left(x-3\right)+xy\left(x-3\right)\)

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

\(=x\left(1+y\right)\left(x-3\right)\)

d) \(2x\left(x-2\right)-6\left(x-2\right)\)

\(=\left(2x-6\right)\left(x-2\right)\)

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