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a) 5x3 - 40 = 5( x3 - 8 ) = 5( x - 2 )( x2 + 2x + 4 )
b) x2z + 4xyz + 4y2z = z( x2 + 4xy + 4y2 ) = z( x + 2y )2
c) 4x2 - y2 - 6x + 3y = ( 4x2 - y2 ) - ( 6x - 3y ) = ( 2x - y )( 2x + y ) - 3( 2x - y ) = ( 2x - y )( 2x + y - 3 )
d) x2 + 2x - 4y2 + 1 = ( x2 + 2x + 1 ) - 4y2 = ( x + 1 )2 - ( 2y )2 = ( x - 2y + 1 )( x + 2y + 1 )
e) 3x2 - 3y2 - 12x + 12y = 3( x2 - y2 - 4x + 4y ) = 3[ ( x2 - y2 ) - ( 4x - 4y ) ] = 3[ ( x - y )( x + y ) - 4( x - y ) ] = 3( x - y )( x + y - 4 )
f) x3 + 5x2 + 4x + 20 = x2( x + 5 ) + 4( x + 5 ) = ( x + 5 )( x2 + 4 )
g) x3 - x2 - 25x + 25 = x2( x - 1 ) - 25( x - 1 ) = ( x - 1 )( x2 - 25 ) = ( x - 1 )( x - 5 )( x + 5 )
a) \(5x^3-40=5\left(x^3-8\right)=5\left(x-2\right)\left(x^2+2x+4\right)\)
b) \(x^2z+4xyz+4y^2z=z\left(x^2+4xy+4y^2\right)=z\left(x+2y\right)^2\)
c) \(4x^2-y^2-6x+3y=\left(4x^2-y^2\right)-\left(6x-3y\right)\)
\(=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)
d) \(x^2+2x-4y^2+1=x^2+2x+1-4y^2\)
\(=\left(x+1\right)^2-4y^2=\left(x+2y+1\right)\left(x-2y+1\right)\)
e) \(3x^2-3y^2-12x+12y=3\left(x^2-y^2-4x+4y\right)\)
\(=3\left[\left(x^2-y^2\right)-\left(4x-4y\right)\right]=3\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]\)
\(=3\left(x-y\right)\left(x+y+4\right)\)
f) \(x^3+5x^2+4x+20=\left(x^3+5x^2\right)+\left(4x+20\right)\)
\(=x^2.\left(x+5\right)+4\left(x+5\right)=\left(x^2+4\right)\left(x+5\right)\)
g) \(x^3-x^2-25x+25=\left(x^3-x^2\right)-\left(25x-25\right)\)
\(=x^2\left(x-1\right)-25\left(x-1\right)=\left(x-1\right)\left(x^2-25\right)\)
\(=\left(x-1\right)\left(x-5\right)\left(x+5\right)\)
Phân tích các đa thức sau thành nhân tử:
a) x(y2-z2)+y(z2-x2)+z(x2-y2)
b) x(y+z)2+y(z+x)2+z(x+y)2-4xyz
b)x(y+z)2+y(z+x)2+z(x+y)2-4xyz
=[x(y+z)2-2xyz]+[y(z+x)2-2xyz]+z(x+y)2
=x(y2+2yz+z2-2yz)+y(x2+z2+2xz-2xz)+z(x+y)2
=x(y2+z2)+y(x2+z2)+z(x+y)2
=xy2+xz2+x2y+yz2+(xz+yz)(x+y)
=xy(x+y)+z2(x+y)+(xz+yz)(x+y)
=(x+y)(xy+z2+xz+yz)
=(x+y)[x(y+z)+z(y+z)]
=(x+y)(y+z)(x+z)
a)x(y2-z2)+y(z2-x2)+z(x2-y2)
=x(y-z)(y+z)+yz2-x2y+x2z-y2z
=(y-z)(xy+xz)-x2(y-z)-yz(y-z)
=(y-z)(xy+xz-x2-yz)
=(y-z)[x(y-x)-z(y-x)]
=(y-z)(y-x)(x-z)
viết lại cho đối xưng thì thế này
\(\left(-x+y+z\right)\left(x-y+z\right)\left(x+y-z\right)\)
x(y+z)^2 - y(z-x)^2 +z(x+y)^2 - x^3 + y^3 - z^3 - 4xyz
=xy^2+2xyz+xz^2-yz^2+2xyz-x^2y+x^2z+2xyz+zy^2-x^3+y^3-z^3-4xyz
=xy^2+xz^2-yz^2-x^2y+x^2z+y^2z-x^3+y^3-z^3+2xyz
=(xy^2+2xyz+xz^2)-x^3-(yz^2+2xyz+x^2y)+y^3+(x^2z+2xyz+y^2z)-z^3
=x[(y+z)^2-x^2)-y[(z+x)^2-y^2]+z[(x+y)^2-z^2]
=x(-x+y+z)(x+y+z)-y(x-y+z)(x+y+z)+z(x+y-z)(x+y+z)
=(x+y+z)[-x^2+xy+xz-xy+y^2-yz+xz+yz-z^2]
=(x+y+z)[-x(x-y-z)-y(x-y-z)+z(x-y-z)]
=(x+y+z)(x-y-z)(z-x-y)
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) \(2x^2-50\)
\(=2\left(x^2-25\right)\)
\(=2\left(x-5\right)\left(x+5\right)\)
b) \(x^2z+4xyz+4y^2z\)
\(=z\left(x^2+4xy+4y^2\right)\)
\(=z\left(x+2y\right)^2\)
c) \(x^2-y^2+12y-36\)
\(=x^2-\left(y^2-2\cdot x\cdot6+6^2\right)\)
\(=x^2-\left(y-6\right)^2\)
\(=\left(x-y+6\right)\left(x+y-6\right)\)
d) Đặt \(D=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(D=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]+1\)
\(D=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(x^2+3x+1=a\)
\(D=\left(a-1\right)\left(a+1\right)+1\)
\(D=a^2-1^2+1\)
\(D=a^2\)
Thay \(x^2+3x+1=a\)vào D ta có :
\(D=\left(x^2+3x+1\right)^2\)
a: \(2x^2-50=2\left(x^2-5^2\right)\)
\(=2\left(x-5\right)\left(x+5\right)\)
b:\(x^2z+4xyz+4y^2z=z\left(x^2+4xy+4y^2\right)\)
\(=z\left(x+2y\right)^2\)
c:\(x^2-y^2+12y-36=x^2-\left(y^2-12y+36\right)\)
\(=x^2-\left(y-6\right)^2\)
\(=\left(x-y+6\right)\left(x+y-6\right)\)
d:\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(t=x^2+3x+1\)Ta có:
\(=\left(t-1\right)\left(t+1\right)+1\)
\(=t^2-1+1=t^2\)
\(=\left(x^2+3x+1\right)^2\)