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xy(x+y)+yz(y+z)+xz(x+z)+2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(y + x)
= (x + y)(xy + zx + zy + z²)
= (x + y)[x(y + z) + z(y + z)]
= (x + y)(y + z)(z + x)
xy(x+y)+yz(y+z)+xz(x+z)+2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(y + x)
= (x + y)(xy + zx + zy + z²)
= (x + y)[x(y + z) + z(y + z)]
= (x + y)(y + z)(z + x)
\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xz^2+xyz\right)+\left(y^2z+yz^2+xyz\right)\)
\(=xy\left(x+y+z\right)+xz\left(x+y+z\right)+yz\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(xy+xz+yz\right)\)
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).
\(a) x^2 - xy+x-y\) \(= (x^2 - xy) + ( x- y) \)
\(=x(x-y) + (x-y)\)
\(= (x-y) (x+1)\)
\(b) xz + yz - 5(x+y)\) \(= (xz + yz) - 5(x+y)\)
\(= z(x+y) - 5(x+y)\)
\(= (x+y) (z-5)\)
\(c) 3x^2 - 3xy - 5x +5y = (3x^2-3xy) - (5x-5y)\)
\(= 3x(x-y) - 5(x-y)\)
\(= (x-y)(3x-5)\)
\(x^2-2xy+y^2-xz+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz.\)
\(=x^2.\left(y+z\right)+yz.\left(y+z\right)+x\left(y^2+z^3\right)+2xyz\)
\(=\left(y+z\right).\left(x^2+yz\right)+x\left(y^{^2}+z^2+2yz\right)\)
\(=\left(y+z\right).\left[x.\left(x+2\right)+y.\left(x+2\right)\right]\)
\(=\left(y+z\right).\left(x+z\right).\left(x+y\right)\)
xy(x+y)+yz(y+z)+xz(x+z)+2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(y + x)
= (x + y)(xy + zx + zy + z2)
= (x + y)[x(y + z) + z(y + z)]
= (x + y)(y + z)(z + x)
\(xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)
\(=\left(xyz-xy-xz+x\right)-yz+y+z-1\)
\(=x\left(yz-y-z+1\right)-\left(yz-y-z+1\right)\)
\(=\left(x-1\right)\left(yz-y-z+1\right)\)
\(=\left(x-1\right)\left[y\left(z-1\right)-\left(z-1\right)\right]\)
\(=\left(x-1\right)\left(y-1\right)\left(z-1\right)\)
xz + yz – 5(x + y)
= (xz + yz) – 5(x + y)
(Nhóm thứ nhất có nhân tử chung là z ; nhóm thứ hai có nhân tử chung là 5)
= z(x + y) – 5(x + y)
(Xuất hiện nhân tử chung là x + y)
= (z – 5)(x + y)