PK
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
PT
1
26 tháng 10 2020
b) = (x^2)^2 + 18x^2 + 9^2 -18x^2
= (x^2 + 9) ^2 - 18x^2
= ( x^2 + 9 - 18x ) ( x^2 +9 + 18x)
PK
0
PK
0
SK
0
NN
14 tháng 10 2020
1) \(4x^2-7x-2=4x^2-8x+x-2=\left(4x^2-8x\right)+\left(x-2\right)\)
\(=4x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(4x+1\right)\)
2) \(4x^2+5x-6=4x^2+8x-3x-6=\left(4x^2+8x\right)-\left(3x+6\right)\)
\(=4x\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(4x-3\right)\)
3) \(5x^2-18x-8=5x^2-20x+2x-8=\left(5x^2-20x\right)+\left(2x-8\right)\)
\(=5x\left(x-4\right)+2\left(x-4\right)=\left(x-4\right)\left(5x+2\right)\)
4) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
\(=xy\left(x+y\right)-y^2z-yz^2+x^2z-xz^2\)
\(=xy\left(x+y\right)+\left(x^2z-y^2z\right)-\left(yz^2+xz^2\right)\)
\(=xy\left(x+y\right)+z\left(x^2-y^2\right)-z^2.\left(x+y\right)\)
\(=xy\left(x+y\right)+z\left(x-y\right)\left(x+y\right)-z^2\left(x+y\right)\)
\(=xy\left(x+y\right)+\left(zx-zy\right)\left(x+y\right)-z^2\left(x+y\right)\)
\(=\left(x+y\right)\left(xy+xz-yz-z^2\right)=\left(x+y\right).\left[x\left(y+z\right)-z\left(y+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(x-z\right)\)
14 tháng 10 2020
1) 4x2 - 7x - 2 = 4x2 - 8x + x - 2 = 4x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 4x + 1 )
2) 4x2 + 5x - 6 = 4x2 - 8x + 3x - 6 = 4x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 4x + 3 )
3) 5x2 - 18x - 8 = 5x2 - 20x + 2x - 8 = 5x( x - 4 ) + 2( x - 4 ) = ( x - 4 )( 5x + 2 )
4) xy( x + y ) - yz( y + z ) + xz( x - z )
= x2y + xy2 - y2z - yz2 + xz( x - z )
= ( x2y - yz2 ) + ( xy2 - y2z ) + xz( x - z )
= y( x2 - z2 ) + y2( x - z ) + xz( x - z )
= y( x - z )( x + z ) + y2( x - z ) + xz( x - z )
= ( x - z )[ y( x + z ) + y2 + xz ]
= ( x - z )( xy + yz + y2 + xz )
= ( x - z )[ ( xy + y2 ) + ( xz + yz ) ]
= ( x - z )[ y( x + y ) + z( x + y ) ]
= ( x - z )( x + y )( y + z )
5) xy( x + y ) + yz + xz( x + z ) + 2xyz ( đề có thiếu không vậy .-. )
TM
25 tháng 9 2017
a)\(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
b)\(x^2+11x+24=x^2+3x+8x+24=x\left(x+3\right)+8\left(x+3\right)=\left(x+3\right)\left(x+8\right)\)
c)\(xy\left(x-y\right)+y^2\left(y-z\right)+zx\left(z-x\right)=x\left[y\left(x-y\right)+z\left(z-x\right)\right]+y^2\left(y-z\right)\)
\(=x\left(xy-y^2+z^2-xz\right)+y^2\left(y-z\right)\)\(=x\left[\left(xy-xz\right)-\left(y^2-z^2\right)\right]+y^2\left(y-z\right)\)
\(=x\left[x\left(y-z\right)-\left(y-z\right)\left(y+z\right)\right]+y^2\left(y-z\right)\)\(=x\left(y-z\right)\left(x-y-z\right)+y^2\left(y-z\right)\)
\(=\left(y-z\right)\left(x^2-xy-xz\right)+y^2\left(y-z\right)=\left(y-z\right)\left(x^2-xy-xz+y^2\right)\)
NP
0
DC
2
25 tháng 10 2020
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-8\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-8\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-8\)
\(=\left(x^2+5x+5\right)^2-1-8\)
\(=\left(x^2+5x+5\right)^2-3^2\)
\(=\left(x^2+5x+2\right)\left(x^2+5x+8\right)\)
b) \(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)
\(=xy\left(x-y\right)+y^2z-yz^2+z^2x-zx^2\)
\(=xy\left(x-y\right)+z^2\left(x-y\right)-z\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(xy+z^2-zx-yz\right)\)
\(=\left(x-y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]\)
\(=\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
25 tháng 10 2020
a) ( x + 1 )( x + 2 )( x + 3 )( x + 4 ) - 8
= [ ( x + 1 )( x + 4 ) ][ ( x + 2 )( x + 3 ) ] - 8
= ( x2 + 5x + 4 )( x2 + 5x + 6 ) - 8
Đặt t = x2 + 5x + 5
bthuc ⇔ ( t - 1 )( t + 1 ) - 8
= t2 - 1 - 8
= t2 - 9
= ( t - 3 )( t + 3 )
= ( x2 + 5x + 5 - 3 )( x2 + 5x + 5 + 3 )
= ( x2 + 5x + 2 )( x2 + 5x + 8 )
b) xy( x - y ) + yz( y - z ) + zx( z - x )
= x2y - xy2 + y2z - yz2 + zx( z - x )
= ( y2z - xy2 ) - ( yz2 - x2y ) + zx( z - x )
= y2( z - x ) - y( z2 - x2 ) + zx( z - x )
= ( z - x )( y2 + zx ) - y( z - x )( z + x )
= ( z - x )( y2 + zx - yz - yx )
= ( z - x )[ ( y2 - yx ) - ( yz - zx ) ]
= ( z - x )[ y( y - x ) - z( y - x ) ]
= ( z - x )( y - x )( y - z )
NM
0