mk

a) (x²-6xy+9y²):(3y-x)

= (x-3y)²:(3y-x)

=(3y-x)²:(3y-x)

= 3y-x

b) (8x³-1):(4x²+2x+1)

=[(2x)³-1]:(4x²+2x+1)

= (2x-1)(4x²+2x+1):(4x²+2x+1)

= 2x-1

(125x³+1):(5x+1)

= [(5x)³+1)]:(5x+1)

=(5x+1)(25x²-5x+1):(5x+1)

= 25x²-5x+1

c1 :

2,38 x 2,04 : 0,68

= 4,8552 : 0,68

= 7,14

C2

2,38 x 2,04 : 0,68

= 2,38 x 3

= 7,14

C3

2,38 x 2,04 : 0,68

= ( 2,38 : 0,68 ) x 2,04

= 3,5 x 2,38

= 7 , 14

b)B=x³-2x-1

= x³+0x²-2x-1

=x³+x²-x²-x-x-1

=(x³+x²)-(x²+x)-(x+1)

=x²(x+1)-x(x+1)-(x+1)

=(x+1)(x²-x-1)

C=x⁵-x⁴+x³-x²+x-1

=(x⁵-x⁴)+(x³-x²)+(x-1)

=x⁴(x-1)+x²(x-1)+(x-1)

=(x-1)(x⁴+x²+1)

a)A=x³+x+2

= x³+0x²+x+2

= x³+x²-x²-x+2x+2

= (x³+x²)-(x²+x)+(2x+2)

=x²(x+1)-x(x+1)+2(x+1)

=(x+1)(x²-x+2)

b) \(\left(2n+7\right)⋮\left(n+1\right)\)

vì\(\left(n+1\right)⋮\left(n+1\right)\)

=>\(2\left(n+1\right)⋮\left(n+1\right)\)

=> \(\left(2n+2\right)⋮\left(n+1\right)\)

=>\(\left(2n+7\right)-\left(2n+2\right)⋮\left(n+1\right)\)

=>\(\left(2n+7-2n-2\right)⋮\left(n+1\right)\)

=>\(5⋮\left(n+1\right)\)

=> \(\left(n+1\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

TA CÓ BẢNG

vậy \(n\in\left\{0;4\right\}\)