Phân tích các đa thức sau thành nhân tử
a) \(x^2-xy+x-y\)
b) \(xz+yz-5\left(x+y\right)\)
c) \(3x^2-3xy-5x+5y\)
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a) x2 – xy + x – y = (x2 – xy) + (x – y)
= x(x – y) + (x -y)
= (x – y)(x + 1)
b) xz + yz – 5(x + y) = z(x + y) – 5(x + y)
= (x + y)(z – 5)
c) 3x2 – 3xy – 5x + 5y = (3x2 – 3xy) – (5x – 5y)
= 3x(x – y) -5(x – y) = (x – y)(3x – 5).
Học tốt nhé !!! Rất hân hạnh ^_^
a) x2 – xy + x – y = (x2 – xy) + (x - y)
= x(x - y) + (x -y)
= (x - y)(x + 1)
b) xz + yz – 5(x + y) = z(x + y) - 5(x + y)
= (x + y)(z - 5)
c) 3x2 – 3xy – 5x + 5y = (3x2 – 3xy) - (5x - 5y)
= 3x(x - y) -5(x - y) = (x - y)(3x - 5).
a.\(xz+yz-5\left(x+y\right)\)
\(=z\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(z-5\right)\)
b.\(3x^2-3xy-5x+5y\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
c.\(x^2+6x-y^2-3z^2\)???Sai đề bài ...?
d.\(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-z^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)'
\(=3\left(x+y-z\right)\left(x+y+z\right)\)
Trả lời:
a, xz + yz - 5 ( x + y )
= ( xz + yz ) - 5 ( x + y )
= z ( x + y ) - 5 ( x + y )
= ( x + y ) ( z - 5 )
b, 3x2 - 3xy - 5x + 5y
= ( 3x2 - 3xy ) - ( 5x - 5y )
= 3x ( x - y ) - 5 ( x - y )
= ( x - y ) ( 3x - 5 )
c, x2 + 6x - y2 - 3z2
= - ( 3x2 - x2 + y2 - 6x )
d, 3x2 + 6xy + 3y2 - 3z2
= 3 ( x2 + 2xy + y2 - x2 )
= 3 [ ( x2 + 2xy + y2 ) - z2 ]
= 3 [ ( x + y )2 - z2 ]
= 3 ( x + y - z ) ( x + y + z )
a)x2-xy+x-y
=x(x-y)+(x-y)
=(x+1)(x-y)
b)3x2-3xy-5x+5y
=3x(x-y)-5(x-y)
=(3x-5)(x-y)
a ) \(x^2-xy+x-y\).
\(=x\left(x-y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+1\right).\)
b ) \(3x^2-3xy-5x+5y\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
a) xy(x + y) + yz(y + z) + xz(z + x) + 3xyz
= xy(X + y + z) + yz(x + y + z) + xz(X + y + z)
= (x + y +z)(xy + yz+ xz)
b) xy(x + y) - yz(y + z) - xz(z - x)
= x2y + xy2 - y2z - yz2 - xz2 + x2z
= x2(y + z) - yz(y + z) + x(y2 - z2)
= x2(y + z) - yz(y + z) + x(y + z)(y - z)
= (y + z)(x2 - yz + xy - xz)
= (y + z)[x(x + y) - z(x + y)]
= (y + z)(x + y)(x - z)
c) x(y2 - z2) + y(z2 - x2) + z(x2 - y2)
= x(y - z)(y + z) + yz2 - yx2 + x2z - y2z
= x(y - z)(y + z) - yz(y - z) - x2(y - z)
= (y - z)((xy + xz - yz - x2)
= (y - z)[x(y - x) - z(y - x)]
= (y - z)(x - z)(y -x)
\(x^2-xy+x-y\)
\(=x\left(x-y\right)+\left(x-y\right)\)
\(=\left(x-1\right)\left(x-y\right)\)
a, \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right).\)
\(b,3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
13: =x(a^3-1)-b(a-1)
=x(a-1)(a^2+a+1)-b(a-1)
=(a-1)(a^2x+a*x+x-b)
12: =(x-y)(x+y)-(x-y)
=(x-y)(x+y-1)
10: =3(x^2-4y^2)
=3(x-2y)*(x+2y)
7: =x^2-x-5x+5=(x-1)(x-5)
8: =x^2+3x+4x+12=(x+3)(x+4)
9: =2x^2-6x-x+3=(x-3)(2x-1)
Đặt x^2+y^2+z^2 =a ; xy+yz+zx=b
=> (x+y+z)^2 =x^2+y^2+z^2+2xy+2yz+2zx =a+2b
Ta có A= (x^2+y^2+z^2)(xy+yz+zx) +(x+y+z)^2
= a(a+2b)+b^2=a^2+2ab+b^2=(a+b)^2
=(x^2+y^2+z^2 +xy+yz+zx)^2
Bài giải:
a) x2 – xy + x – y = (x2 – xy) + (x - y)
= x(x - y) + (x -y)
= (x - y)(x + 1)
b) xz + yz – 5(x + y) = z(x + y) - 5(x + y)
= (x + y)(z - 5)
c) 3x2 – 3xy – 5x + 5y = (3x2 – 3xy) - (5x - 5y)
= 3x(x - y) -5(x - y) = (x - y)(3x - 5).
\(a) x^2 - xy+x-y\) \(= (x^2 - xy) + ( x- y) \)
\(=x(x-y) + (x-y)\)
\(= (x-y) (x+1)\)
\(b) xz + yz - 5(x+y)\) \(= (xz + yz) - 5(x+y)\)
\(= z(x+y) - 5(x+y)\)
\(= (x+y) (z-5)\)
\(c) 3x^2 - 3xy - 5x +5y = (3x^2-3xy) - (5x-5y)\)
\(= 3x(x-y) - 5(x-y)\)
\(= (x-y)(3x-5)\)