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a.\(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
b.\(x^3-6x^2+12x-8=\left(x-2\right)^3\)
c.\(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)
a)\(6x-9-x^2\)
\(=-\left(x^2+6x+9\right)\)
\(=-\left(x+3\right)^2\)
b)\(x^2+4y^2+4xy\)
\(=\left(x+2y\right)^2\)
c)\(x^2+8x+16\)
\(=\left(x+4\right)^2\)
d)\(9x^2-12xy+4y^2\)
\(=\left(3x-2y\right)^2\)
e)\(-25x^2y^2+10xy-1\)
\(=-\left(25x^2y^2-10xy+1\right)\)
\(=-\left(5xy-1\right)^2\)
f)\(4x^2-4x+1\)
\(=\left(2x-1\right)^2\)
j)\(x^2+6x+9\)
\(=\left(x+3\right)^2\)
h)\(9x^2-6x+1\)
\(=\left(3x-1\right)^2\)
#H
a, 6x - 9 - x2 = - x2 + 6x - 9 = - (x2 - 6x + 9) = - (x - 3)2
b, x2 + 4y2 + 4xy = x2 + 2. x . 2y + (2y)2 = (x + 2y)2
c, x2 + 8x + 16 = x2 + 2 . x . 4 + 42 = (x + 4)2
d, 9x2 - 12xy + 4y2 = (3x)2 - 2 . 3x . 2y + (2y)2 = (3x - 2y)2
e, - 25x2y2 + 10xy - 1 = - (25x2y2 - 10xy + 1) = - [(5xy)2 - 2 . 5xy + 1] = - (5xy - 1)2
f, 4x2 - 4x + 1 = (2x)2 - 2 . 2x + 1 = (2x - 1)2
j, x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
h, 9x2 - 6x + 1 = (3x)2 - 2 . 3x + 1 = (3x - 1)2
1/ \(4x^2-12xy+9y^2=\left(2x\right)^2-2.2.3xy+\left(3y\right)^2\)
\(=\left(2x-3y\right)^2\)
2/ \(x^3-y^6=x^3-\left(y^2\right)^3\)
\(=\left(x-y^2\right)\left(x^2+xy^2+y^4\right)\)
Làm tạm 2 phần đợi mik xíu
4x2 - 12xy + 9y2 = ( 2x )2 - 2.2x.3y + ( 3y )2 = ( 2x - 3y )2
x3 - y6 = x3 - ( y2 )3 = ( x - y2 )( x2 + xy2 + y4 )
x6 - 6x4 + 12x2 - 8 = ( x2 )3 - 3.(x2)2.2 + 3.x2.22 - 23 = ( x2 - 2 )3
( x2 + 4y2 - 5 )2 - 16( x2y2 + 2xy + 1 ) = ( x2 + 4y2 - 5 )2 - 42( xy + 1 )2
= ( x2 + 4y2 - 5 )2 - ( 4xy + 4 )2
= [ ( x2 + 4y2 - 5 ) - ( 4xy + 4 ) ][ ( x2 + 4y2 - 5 ) + ( 4xy + 4 ) ]
= ( x2 + 4y2 - 5 - 4xy - 4 )( x2 + 4y2 - 5 + 4xy + 4 )
= [ ( x2 - 4xy + 4y2 ) - 9 ][ ( x2 + 4xy + 4y2 ) - 1 ]
= [ ( x - 2y )2 - 32 ][ ( x + 2y )2 - 12 ]
= ( x - 2y - 3 )( x - 2y + 3 )( x + 2y - 1 )( x + 2y + 1 )
( a + b )3 - ( a3 + b3 ) = a3 + 3a2b + 3ab2 + b3 - a3 - b3
= 3a2b + 3ab2
= 3ab( a + b )
\(=9x^3-3x^2-9x^2+6x-1\)1
\(=3x^2\left(3x-1\right)-\left(9x^2-6x+1\right)\)
\(=3x^2\left(3x-1\right)-\left(3x-1\right)^2\)
\(=\left(3x-1\right)\left(3x^2-3x+1\right)\)
2.\(\left(x^2-16y^2\right)-3x+12y=\left(x-4y\right)\left(x+4y\right)-3\left(x-4y\right)=\left(x-4y\right)\left(x+4y-3\right)\)
4.\(x^3+6x^2+12x+8=\left(x+2\right)^3\)
Trả lời:
j, ( x + 1 )2 - ( 2x - 1 )2 = 0
<=> ( x + 1 - 2x + 1 ) ( x + 1 + 2x - 1 ) = 0
<=> ( 2 - x ) 3x = 0
<=> 2 - x = 0 hoặc 3x = 0
<=> x = 2 hoặc x = 0
Vậy x = 2; x = 0 là nghiệm của pt.
k, Sửa đề: 8x3 + 12x - 1 = 6x2
<=> 8x3 + 12x - 1 - 6x2 = 0
<=> ( 2x )2 - 3.x2.2 + 3.x.22 - 13 = 0
<=> ( 2x - 1 )3 = 0
<=> 2x - 1 = 0
<=> 2x = 1
<=> x = 1/2
Vậy x = 1/2 là nghiệm của pt.
l, x3 + 15x2 + 75x + 125 = 0
<=> x3 + 3.x2.5 + 3.x.52 + 53 = 0
<=> ( x + 5 )3 = 0
<=> x + 5 = 0
<=> x = - 5
Vậy x = - 5 là nghiệm của pt.
A = ( 3x )3 + 23 - 27x3 + 6 = 27x3 + 8 - 27x3 + 6 = 14 ( đpcm )
B = x3 + 3x2 + 3x + 1 - ( x3 - 1 ) - 3x2 - 3x = x3 + 1 - x3 + 1 = 2 ( đpcm )
C = 6( x + 2 )( x2 - 2x )( x2 - 2x + 4 ) - 6x3 - 2 ( bạn xem lại đề bài nhé ._. )
D = 2[ ( 3x )3 + 13 ] - 54x3 = 2( 27x3 + 1 ) - 54x3 = 54x3 + 2 - 54x3 = 2 ( đpcm )