Question

[Khánh Linh Đỗ]

giúp mình câu 13 14 15

Answer 1

[Nguyễn Lê Phước Thịnh]

Câu 13:

1: 

a: \(2x^2+2x=2x\cdot x+2x\cdot1=2x\left(x+1\right)\)

b: \(9x^2-4y^2\)

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

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

2:

\(\dfrac{xy+2x+1}{xy+x+y+1}+\dfrac{yz+2y+1}{yz+y+z+1}+\dfrac{zx+2z+1}{zx+z+x+1}\)

\(=\dfrac{xy+2x+1}{\left(y+1\right)\left(x+1\right)}+\dfrac{yz+2y+1}{\left(z+1\right)\left(y+1\right)}+\dfrac{z\left(x+2\right)+1}{\left(z+1\right)\left(x+1\right)}\)

\(=\dfrac{\left(xy+2x+1\right)\left(z+1\right)+\left(yz+2y+1\right)\left(x+1\right)+\left(xz+2z+1\right)\left(y+1\right)}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(=\dfrac{xyz+xy+2xz+2x+z+1+xyz+yz+2xy+2y+x+1+\left(xz+2z+1\right)\left(y+1\right)}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(=\dfrac{2xyz+3xy+2xz+3x+z+2+yz+2y+x+xyz+xz+2zy+2z+y+1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(=\dfrac{3xyz+3xy+3xz+3yz+3x+3z+3y+3}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(=\dfrac{3\left(xyz+xy+xz+yz+x+z+y+1\right)}{\left(xy+x+y+1\right)\left(z+1\right)}\)

=3

Câu 14:

1:

f(0)=0+5=5

2:
Vì hệ số góc của y=ax+b là -1 nên a=-1

=>y=-x+b

Thay x=1 và y=2 vào y=-x+b, ta được:

b-1=2

=>b=3