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6 tháng 7 2016
\(\frac{27x^3-1}{3x-1}=\frac{\left(3x\right)^3-1}{3x-1}=\frac{\left(3x-1\right)\left(9x^2+3x+1\right)}{3x-1}=9x^2+3x+1\)
ủng hộ 1 t i c k nha cảm ơn
H
0
21 tháng 7 2016
a ) \(\left(4x^2-9y^2\right):\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y\right);\left(2x-3y\right)\)
\(=2x+3y\)
b ) \(\left(27x^3-1\right):\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right):\left(3x-1\right)\)
\(=9x^2+3x+1\)
21 tháng 7 2016
a) (4x2 – 9y2) : (2x – 3y);
=(2x-3y)(2x+3y):(2x-3y)
=2x+3y
b) (27x3 – 1) : (3x – 1);
=(3x-1)(9x2+3x+1):(3x-1)
=9x2+3x+1
c) (8x3 + 1) : (4x2 – 2x + 1);
=(2x+1)(4x2-2x+1):(4x2-2x+1)
=2x+1
d) (x2 – 3x + xy -3y) : (x + y)
=[x.(x-3)+y.(x-3)]:(x+y)
=(x-3)(x+y):(x+y)
=x-3
MT
6 tháng 8 2015
a) (4x2 – 9y2) : (2x – 3y);
=(2x-3y)(2x+3y):(2x-3y)
=2x+3y
b) (27x3 – 1) : (3x – 1);
=(3x-1)(9x2+3x+1):(3x-1)
=9x2+3x+1
c) (8x3 + 1) : (4x2 – 2x + 1);
=(2x+1)(4x2-2x+1):(4x2-2x+1)
=2x+1
d) (x2 – 3x + xy -3y) : (x + y)
=[x.(x-3)+y.(x-3)]:(x+y)
=(x-3)(x+y):(x+y)
=x-3
6 tháng 8 2015
a) ( 4x^2 - 9y^2) : ( 2x- 3y) = ( 2x - 3y)(2x + 3y) : (2x- 3y) = 2x + 3y
b) (27x^3 - 1 ) : (3x- 1) = ( 3x - 1 )(9x^2 + 3x + 1 ) : ( 3x - 1 ) = 9x^2 + 3x+ 1
c) ( 9x^3 + 1 ) : (4x^2 - 2x + 1 ) = ( 2x + 1 )(4x^2 - 2x + 1 ) : (4x^2 - 2x + 1 ) = 2x+ 1
d) ( x^2 - 3x + x y - 3y ) : ( x+ y)
= [ x(x- 3 ) + y ( x - 3 ) ] : ( x+ y)
= ( x+ y)(x- 3 ) : ( x+ y)
=x - 3
MT
7 tháng 10 2019
a) \(x^3+6x^2+12x+8\)
\(=\left(x+2\right)^3\)
b) \(x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
c) \(1-9x+27x^2-27x^3\)
\(=-\left(27x^3-27x^2+9x-1\right)\)
\(=-\left(3x-1\right)^3\)
28 tháng 6 2016
a) (4x2 – 9y2) : (2x – 3y) = [(2x)2 – (3y)2] : (2x – 3y) = 2x + 3y;
b) (27x3 – 1) : (3x – 1) = [(3x)3 – 1] : (3x – 1) = (3x)2 + 3x + 1 = 9x2 + 3x + 1
c) (8x3 + 1) : (4x2 – 2x + 1) = [(2x)3 + 1] : (4x2 – 2x + 1)
= (2x + 1)[(2x)2 – 2x + 1] : (4x2 – 2x + 1)
= (2x + 1)(4x2 – 2x + 1) : (4x2 – 2x + 1) = 2x + 1
d) (x2 – 3x + xy -3y) : (x + y)
= [(x2 + xy) – (3x + 3y)] : (x + y)
= [x(x + y) – 3(x + y)] : (x + y)
= (x + y)(x – 3) : (x + y)
= x – 3.
TC
20 tháng 12 2017
a) (4x2−9y2):(2x−3y)=[(2x)2−(3y)2]:(2x−3y)(4x2−9y2):(2x−3y)=[(2x)2−(3y)2]:(2x−3y)
=(2x−3y).(2x+3y):(2x−3y)=2x+3y=(2x−3y).(2x+3y):(2x−3y)=2x+3y;
b) (27x3−1):(3x−1)=[(3x)3−13]:(3x−1)(27x3−1):(3x−1)=[(3x)3−13]:(3x−1)
=(3x−1).[(3x)2+3x+1]:(3x−1)=9x2+3x+1=(3x−1).[(3x)2+3x+1]:(3x−1)=9x2+3x+1
c) (8x3+1):(4x2−2x+1)=[(2x)3+13]:(4x2−2x+1)(8x3+1):(4x2−2x+1)=[(2x)3+13]:(4x2−2x+1)
=(2x+1)[(2x)2−2x+1]:(4x2−2x+1)=(2x+1)[(2x)2−2x+1]:(4x2−2x+1)
=(2x+1)(4x2−2x+1):(4x2−2x+1)=2x+1=(2x+1)(4x2−2x+1):(4x2−2x+1)=2x+1
d) (x2−3x+xy−3y):(x+y)(x2−3x+xy−3y):(x+y)
=[(x2+xy)−(3x+3y)]:(x+y)=[x(x+y)−3(x+y)]:(x+y)=(x+y)(x−3):(x+y)=x−3
Làm thế này nha
\(\frac{27x^3-1}{3x-1}=\frac{\left(3x\right)^3-1}{3x-1}=\frac{\left(3x-1\right)\left(9x^2+3x+1\right)}{3x-1}=9x^2+3x+1\)
Ta có: \(\left(27x^3-1\right):\left(3x-1\right)=\left[\left(3x\right)^3-1^3\right]:\left(3x-1\right)=\left(3x-1\right)\left(9x^2+3x+1\right):\left(3x-1\right)=9x^2+3x+1\)