Phân tích các đa thức sau thành nhân tử:
a) (x-y)3+(y-z)3+(z-x)3
b)x(y-z)3+y(z-x)3+z(x-y)3
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Do câu d mình ko biết làm bởi v mình không làm được
`a, 4a^2 + 4a + 1 = (2a+1)^2`
`b, -3x^2 + 6xy - 3y^2`
` = -3(x-y)^2`
`c, (x+y)^2 - 2(x+y)z + z^2`
`= (x+y-z)^2`
`a, P = 2x(3 - x^2)`
`b, Q = 5x^2(x-3y)`
`c, R = xy(3x^2y^2 - 6y^2z + 1)`
a) \(P=6x-2x^3\)
\(P=2x\left(3+x^2\right)\)
b) \(Q=5x^3-15x^2y\)
\(Q=5x^2\left(x-3y\right)\)
c) \(R=3x^3y^3-6xy^3z+xy\)
\(R=xy\left(3x^2y^2-6y^2z+1\right)\)
t chỉ cho kết quả thôi nhá, còn nhóm nhân tử you tự xử nhá !
=(x-y)(z-x)(z-y)(x+y+z)
\(\left(x-y\right)z^3+\left(z-z\right)y^3+\left(y-z\right)x^3\)
\(=z^3\left(x-y\right)+y^3\left(z-x\right)+x^3\left(y-z\right)\)
\(=xz^3-yz^3+\left(z-x\right)y^3+\left(y-z\right)x^3\)
\(=xz^3-yz^3+y^3z-xy^3+\left(y-z\right)x^3\)
\(=xz^3-yz^3+y^3z-xy^3+y^3z-xy^3+x^3y-x^3z\)
Mk ko chắc
1)
a) \(=3x^2\left(x^2-1\right)-\left(x^3-1\right)+x^8-3x^4+3x^2-1\)
\(=3x^4-3x^2-x^3+1+x^8-3x^4+3x^2-1=x^8-x^3\)
2)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)-6\left(x^2+5x\right)+45\)
\(=\left(x^2+5x\right)^2-6\left(x^2+5x\right)-36+45\)
\(=\left(x^2+5x\right)^2-6\left(x^2+5x\right)+9=\left(x^2+5x-3\right)^2\)
-(z+x)3 mới đúng-
đặt x+y=a , y+z=b , z+x=c thì a+b+c=2(x+y+z)
ta có 8(x+y+z)3-(x+y)3-(y+z)3-(z+x)3=[2(x+y+z)]3-(x+y)3-(y+z)3-(z+x)3=(a+b+c)3-a3-b3-c3=3(a+b)(b+c)(c+a)
=3(x+2y+z)(y+2z+x)(z+2x+y)
Đặt \(\left\{{}\begin{matrix}a=x+y\\b=y+z\\c=x+z\end{matrix}\right.\Leftrightarrow x+y+z=\dfrac{a+b+c}{2}\)
\(8\left(x+y+z\right)^3-\left(x+y\right)^3-\left(y+z\right)^3-\left(z+x\right)^3\\ =8\left(\dfrac{a+b+c}{2}\right)^3-a^3-b^3-c^3\\ =\left(a+b+c\right)^3-a^3-b^3-c^3\\ =\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)-\left(a+b\right)^3+3ab\left(a+b\right)-c^3\\ =3\left(a+b\right)\left(ac+bc+c^2+ab\right)\\ =3\left(a+b\right)\left(b+c\right)\left(a+c\right)\\ =3\left(x+y+y+z\right)\left(y+z+z+x\right)\left(z+x+x+y\right)\\ =3\left(x+2y+z\right)\left(x+y+2z\right)\left(2x+y+z\right)\)
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: \(2x^2+3xy-14y^2\)
\(=2x^2+7xy-4xy-14y^2\)
\(=\left(2x^2+7xy\right)-\left(4xy+14y^2\right)\)
\(=x\left(2x+7y\right)-2y\left(2x+7y\right)\)
\(=\left(2x+7y\right)\left(x-2y\right)\)
b: \(\left(x-7\right)\left(x-5\right)\left(x-3\right)\left(x-1\right)+7\)
\(=\left(x-7\right)\left(x-1\right)\left(x-5\right)\left(x-3\right)+7\)
\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)+7\)
\(=\left(x^2-8x\right)^2+15\left(x^2-8x\right)+7\left(x^2-8x\right)+105+7\)
\(=\left(x^2-8x\right)^2+22\left(x^2-8x\right)+112\)
\(=\left(x^2-8x\right)^2+8\left(x^2-8x\right)+14\left(x^2-8x\right)+112\)
\(=\left(x^2-8x\right)\left(x^2-8x+8\right)+14\left(x^2-8x+8\right)\)
\(=\left(x^2-8x+8\right)\left(x^2-8x+14\right)\)
c: \(\left(x-3\right)^2+\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)
\(=\left(x-3\right)^2+2\left(x-3\right)\left(3x-1\right)-\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)
\(=\left(x-3\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]-\left(3x-1\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]\)
\(=\left(x-3+6x-2\right)\left(x-3-3x+1\right)\)
\(=\left(7x-5\right)\left(-2x-2\right)\)
\(=-2\left(x+1\right)\left(7x-5\right)\)
d: \(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)
\(=x^2y-xy^2+y^2z-yz^2+zx\left(z-x\right)\)
\(=\left(x^2y-yz^2\right)-\left(xy^2-y^2z\right)+xz\left(z-x\right)\)
\(=y\left(x^2-z^2\right)-y^2\left(x-z\right)-xz\left(x-z\right)\)
\(=y\cdot\left(x-z\right)\left(x+z\right)-\left(x-z\right)\left(y^2+xz\right)\)
\(=\left(x-z\right)\left(xy+zy-y^2-xz\right)\)
\(=\left(x-z\right)\left[\left(xy-y^2\right)+\left(zy-zx\right)\right]\)
\(=\left(x-z\right)\left[y\cdot\left(x-y\right)-z\left(x-y\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(y-z\right)\)