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a) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=x^3+y^3+z^3+3x^2y+3x^2z+3y^2z+3xy^2+3xz^2+3yz^2+6xyz-x^3-y^3-z^2\)
\(=3x^2y+3xy^2+3x^2z+3xz^2+3y^2z+3yz^2+6xyz\)
\(=3xy\left(x+y\right)+3xz\left(x+z\right)+3yz\left(y+z\right)+6xyz\)
\(=3\left[xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+2xyz\right]\)
\(=3\left[xy\left(x+y\right)+x^2z+xz^2+y^2z+yz^2+2xyz\right]\)
\(=3\left[xy\left(x+y\right)+xz\left(x+y\right)+z^2\left(x+y\right)+yz\left(x+y\right)\right]\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
b) \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3\)
\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^3\)
\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(x-z\right)^2\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-y-y+z\right)+\left(y-z-x+z\right)\left(y-z+x-z\right)\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-2y+z\right)-\left(x-y\right)\left(y-2z+x\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(x-2y+z-y+2z-x\right)\)
\(=\left(x-z\right)\left(x-y\right)\left(3z-3y\right)\)
\(=3\left(x-z\right)\left(x-y\right)\left(z-y\right)\)
a) (x + y + z)3 - x3 - y3 - z3
= (x + y + z)3 - z3 - (x3 + y3)
= (x + y + z - z)[(x + y + z)2 + (x + y + z).z + z2) - (x + y)(x2 - xy + y2)
= (x + y)(x2 + y2 + z2 + 2xy + 2yz + 2zx + 2xz + 2yz + z2 + z2) - (x + y)(x2 - xy + y2)
= (x + y)(x2 + y2 + 3z2 + 2xy + 4yz + 4zx) - (x + y)(x2 - xy + y2)
= (x + y)(3z2 + 3xy + 5yz + 4zx)
b) Sửa đề x4 + 2010x2 + 2009x + 2010
= (x4 + x2 + 1) + (2009x2 + 2009x + 2009)
= (x4 + 2x2 + 1 - x2) + 2009(x2 + x + 1)
= [(x2 + 1)2 - x2] + 2009(x2 + x + 1)
= (x2 + x + 1)(x2 - x + 1) + 2009(x2 + x + 1)
= (x2 + x + 1)(x2 - x + 2010)
Ta có :
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y+z\right)^3-x^3\right]-\left(y^3+z^3\right)\)
\(=\left(x+y+z-x\right)\left[\left(x+y+z\right)^2+x^2+\left(x+y+z\right)x\right]-\left(y+z\right)\left(y^2+z^2-yz\right)\)
\(=\left(y+z\right)\left[x^2+y^2+z^2+2\left(xy+yz+xz\right)+x^2+x^2+xy+zx\right]\)\(-\left(y+z\right)\left(y^2+z^2-yz\right)\)
\(=\left(y+z\right)\left[3x^2+y^2+z^2+3xy+3zx+2yz-y^2-z^2+yz\right]\)
\(=\left(y+z\right)\left[3x^2+3xy+3zx+3yz\right]\)
\(=\left(y+z\right)3\left[\left(x^2+xy\right)+\left(zx+yz\right)\right]\)
\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)
Vậy ...
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=x^3+y^3+z^3+2xy+2xz+2yz-x^3-y^3-z^3\)
\(=2xy+2xz+2yz\)
\(=2\left(xy+xz+yz\right)\)
Đc chưa ?
Phương Đỗ Sai rùi bạn.
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y+z\right)\left(x+y\right)z+z^3-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+3\left(x+y+z\right)\left(x+y\right)z+z^3-x^3-y^3-z^3\)
\(=3xy\left(x+y\right)+3\left(x+y+z\right)\left(x+y\right)z\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
4x^4+y^4= 4x^4 +4x^2y^2+y^4-4x^2y^2= ( 2x^2 + y^2 ) ^2 - ( 2xy ) ^2
= (2x^2 + 2xy +y^2)( 2x^2 - 2xy + y^2)
a) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3-\left(y^2+z^2\right)^3\)
\(=\left[\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3\right]-\left(y^2+z^2\right)^3\)
\(=\left(x^2+y^2+z^2-x^2\right)\left[\left(x^2+y^2\right)^2-\left(x^2+y^2\right)\left(z^2-x^2\right)+\left(z^2-x^2\right)^2\right]-\left(y^2+z^2\right)^3\)
\(=\left(y^2+z^2\right)\left(x^4+2x^2y^2+y^4-x^2z^2+x^4-y^2z^2+x^2y^2+z^4-2z^2x^2+x^4\right)-\left(y^2+z^2\right)^3\)
\(=\left(y^2+z^2\right)\left[x^4+2x^2y^2+y^4-x^2z^2+x^4-y^2z^2+x^2y^2+z^4-2z^2x^2+x^4-\left(y^2+z^2\right)^2\right]\)
\(=\left(y^2+z^2\right)\left(x^4+2x^2y^2+y^4-x^2z^2+x^4-y^2z^2+x^2y^2+z^4-2z^2x^2+x^4-y^4-2y^2z^2-z^4\right)\)
\(=\left(y^2+z^2\right)\left(3x^4+3x^2y^2-3x^2z^2-3y^2z^2\right)\)
= 3(y2+z2)(x4+x2y2-x2z2-y2z2)
= 3(y2+z2)[x2(x2+y2)-z2(x2+y2)]
= 3(y2+z2)(x2-z2)(x2+y2)
= 3(y2+z2)(x-z)(x+z)(x2+y2)
b) \(\left(x+y\right)^3-x^3-y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)
\(=3x^2y+3xy^2=3xy\left(x+y\right)\)
c) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2.z+3\left(x+y\right).z^2+z^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2.z+3\left(x+y\right).z^2-\left(x^3+y^3\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2+3\left(x+y\right).z+3z^2\right]-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2+3xz+3yz+3z^2-x^2+xy-y^2\right)\)
= (x+y)[3xy+3xz+3yz+3z2 ]
= 3(x+y)(xy+xz+yz+z2)
= 3(x+y)[x(y+z)+z(y+z)]
= 3(x+y)(x+z)(y+z)
a) \(\left(x^2+y^2\right)^3+\left(z^2-x^3\right)-\left(y^2+z^2\right)^3\)
\(=x^6+3x^4y^2+3x^4y^2+y^6+z^2+-x^2+-y^6+-3y^4z+-3y^2z^4+-z^6\)
\(=x^6+3x^4y^2+3x^2y^4+-3y^4z^4+-z^6+-x^2+z^2\)
b) \(\left(x+y\right)^3-x^3-y^3\)
\(=x^3+3x^2y+3xy^2+y^3+-x^3+-y^3\)
\(=\left(x^3+-x^3\right)+\left(3x^2y\right)+\left(3xy^2\right)+\left(y^3+-y^3\right)\)
\(=3x^2y+3xy^2\)
c) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=x^3+3x^2y+3x^2z+3xy^2+6xyz+3xz^2+y^3+3y^2z+3yz^2+z^2-x^3-y^3-z^3\)
\(=3x^2y+3x^2z+3xy^2+3xy^2+6xyz+3xz^2+3y^2z+3yz^2\)
P/s: Ko chắc