Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
b) sửa đề nhé!
\(6x-9-x^2=-\left(x^2-6x+9\right)\)
\(=-\left(x-3\right)^2\)
a) \(=\left(x-2y\right)\left(x^2+5x\right)\)
b) \(=\left(x-1\right)\left(x^2+2x+1\right)=\left(x-1\right)\left(x+1\right)^2\)
c) \(=\left(x^2+1-2x\right)\left(x^2+1+2x\right)\)
\(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)^2\)
d) \(=3\left(x+3\right)-\left(x-3\right)\left(x+3\right)\)
\(=\left(x+3\right)\left(3-x+3\right)\)
\(=\left(x+3\right)\left(6-x\right)\)
e) \(=\left(x^2-\frac{1}{3}x\right)\left(x^2+\frac{1}{3}x\right)\)
f) \(=2x\left(x-y\right)-16\left(x-y\right)\)
\(=2\left(x-y\right)\left(x-8\right)\)
1, x2(x2+2x+1)=x2(x+1)2
2, 2(x2+2x+1-y2)=2(x+1-y)(x+1+y)
3, 16-(x2+2xy+y2)=(4-x-y)(4+x+y)
\(x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
hk tốt
^^
a) x^4 - x^3 - x + 1
= x^3 ( x - 1 ) - ( x- 1 )
= ( x^3 - 1 )(x - 1)
= ( x- 1 )^2 (x^2 + x + 1 )
a)x4-x3-x+1
=x3(x-1)-(x-1)
=(x-1)(x3-1)
=(x-1)(x-1)(x2+x+1)
=(x-1)2(x2+x+1)
b)5x2-4x+20xy-8y
(sai đề)
Áp dụng HĐT a2 - b2 = ( a - b )( a + b )
và tính chất an.bn = ( a.b )n ( với n ∈ N* )
a) ( 3x + 1 )2 - ( x + 1 )2
= [ ( 3x + 1 ) - ( x + 1 ) ][ ( 3x + 1 ) + ( x + 1 ) ]
= ( 3x + 1 - x - 1 )( 3x + 1 + x + 1 )
= 2x( 4x + 2 )
= 2x.2( 2x + 1 )
= 4x( 2x + 1 )
b) ( x + y )2 - ( x - y )2
= [ ( x + y ) - ( x - y ) ][ ( x + y ) + ( x - y ) ]
= ( x + y - x + y )( x + y + x - y )
= 2y.2x = 4xy
c) ( 2xy + 1 )2 - ( 2x + y )2
= [ ( 2xy + 1 ) - ( 2x + y ) ][ ( 2xy + 1 ) + ( 2x + y ) ]
= ( 2xy + 1 - 2x - y )( 2xy + 1 + 2x + y )
= [ ( 2xy - 2x ) - ( y - 1 ) ][ ( 2xy + 2x ) + ( y + 1 ) ]
= [ 2x( y - 1 ) - ( y - 1 ) ][ 2x( y + 1 ) + ( y + 1 ) ]
= ( y - 1 )( 2x - 1 )9 y + 1 )( 2x + 1 )
d) 9( x - y )2 - 4( x + y )2
= 32( x - y )2 - 22( x + y )2
= [ 3( x - y ) ]2 - [ 2( x + y ) ]2
= ( 3x - 3y )2 - ( 2x + 2y )2
= [ ( 3x - 3y ) - ( 2x + 2y ) ][ ( 3x - 3y ) + ( 2x + 2y ) ]
= ( 3x - 3y - 2x - 2y )( 3x - 3y + 2x + 2y )
= ( x - 5y )( 5x - y )
e) ( 3x - 2y )2 - ( 2x - 3y )2
= [ ( 3x - 2y ) - ( 2x - 3y ) ][ ( 3x - 2y ) + ( 2x - 3y ) ]
= ( 3x - 2y - 2x + 3y )( 3x - 2y + 2x - 3y )
= ( x + y )( 5x - 5y )
= ( x + y )5( x - y )
f) ( 4x2 - 4x + 1 ) - ( x + 1 )2
= ( 2x - 1 )2 - ( x + 1 )2
= [ ( 2x - 1 ) - ( x + 1 ) ][ ( 2x - 1 ) + ( x + 1 ) ]
= ( 2x - 1 - x - 1 )( 2x - 1 + x + 1 )
= 3x( x - 2 )
1) Ta có: 2xy - x2 - y2 + 16
= -(x2 - 2xy + y2 - 16)
= -[(x - y)2 - 16]
= -(x - y - 4)(x - y + 4)
2) x3 + 2x2y + xy2 - 9x
= x(x2 + 2xy + y2 - 9)
= x[(x + y)2 - 9]
= x(x + y - 3)(x + y + 3)
3) x4 - 2x2 = x2(x2 - 2)
1. 2xy-x2-y2+16= -(x2-2xy+y2-16) = -(x2-2xy+y2)-16 = -(x-y)2-16= (x+y)2-42= (x+y-4).(x+y+4)
2. x3+2x2y+xy2-9x= (có sai đề không vậy?)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
1,
a, = 2x.(x-2)
b, = (x^2+y^2+2xy)-(2x+2y)
= (x+y)^2-2.(x+y)
= (x+y).(x+y-2)
2,
a,<=> x^2-1-x^2-2x = 3
<=> -2x-1=3
<=> -2x=4
<=> x=4 : (-2) = -2
b, <=>(x^2-4x+4)-7=0
<=>(x-2)^2-7=0
<=> (x-2)^2=7
=> x-2=+-\(\sqrt{7}\)
<=> x=2+-\(\sqrt{7}\)
k mk nha
a, \(2x-4x\)
\(=-2x\)
b, \(x^2+y^2+2xy-2x-2y\)
\(=\left(x+y\right)^2-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-2\right)\)
a, \(\left(x+1\right)\left(x-1\right)-x\left(x+2\right)=3\)
\(\Leftrightarrow x^2-1-x^2-2x=3\)
\(\Leftrightarrow-2x=4\)
\(\Leftrightarrow x=-2\)
b,\(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
1, \(1-x^2+2xy-y^2=1-\left(x-y\right)^2=\left(1-x+y\right)\left(1+x-y\right)\)
2, \(x^3+x^2-x-1=x^2\left(x+1\right)-\left(x+1\right)=\left(x-1\right)\left(x+1\right)^2\)
3, \(x+2a\left(x-y\right)-y=\left(x-y\right)\left(1+2a\right)\)
4, \(4x^2-9+\left(2x+3\right)^2=\left(2x+3\right)\left(2x-3\right)+\left(2x+3\right)=2\left(2x+3\right)\left(x-1\right)\)
1, \(1-x^2+2xy-y^2=1-\left(x-y\right)^2=\left(1-x+y\right)\left(1+x-y\right)\)
2,\(x+2a\left(x-y\right)-y=\left(x-y\right)+2a\left(x-y\right)=\left(x-y\right)\left(1+2a\right)\)
2,\(x^3+x^2-x-1=x^2\left(x+1\right)-\left(x+1\right)=\left(x-1\right)\left(x+1\right)^2\)
3,\(\left(4x^2-9\right)+\left(2x+3\right)^2=\left(2x+3\right)\left(2x-3\right)+\left(2x+3\right)^2=\left(2x+3\right)4x\)