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a.\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)^3+z^3\right]-a^3-b^3-c^3\)
\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
b.\(x^4+2010x^2+2009x+2010\)
\(=\left(x^4-x\right)+\left(2010x^2+2010x+2010\right)\)
=\(x\left(x-1\right)\left(x^2+x+1\right)+2010\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^2-x+2010\right)\)
sửa đề:\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
giải:
\(\left(x+y+z\right)^3-x^3-y^3-z^3=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)-x^3-y^3-z^3\\ =3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
b,W = \(x^4+x^2+1+2009x^2+2009x+2009\)
\(=\left(x^4+2x^2+1\right)-x^2+2009\left(x^2+x+1\right)\)
\(=\left(x^2+1\right)^2-x^2+2009\left(x^2+x+1\right)\)
\(=\left(x^2+1-x\right)\left(x^2+1+x\right)+2009\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2010\right)\)
1)x(x2 - 19 - 30)
2)x(x2 - 7 - 6)
3)x(x2 + 4x - 7 - 10)
( 4 tích mình làm tiếp 3 câu cuối)
Ta có \(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-xz-yz\right)-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\right]=0\)(Nhân hai vế với 2)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
Tới đây bạn xét hai trường hợp nhé :)
(x+y+z)((X+Y)^2-Z(X+Y))-3XY(X+Y+Z)
=(X+Y+Z)(X^2+2XY+Y^2-XZ-YZ-3XY)
=(X+Y+Z)(X^2+Y^2+Z^2-XZ-YZ-XY)
a) (x+y+z)3 -x3 - y3 - z3
= [(x + y + z) - z][(x+ y + z)2 + x2 + x(x+ y + z)] - (y + z)(y2+ z2 - yz)
= (y+z)(x2 + y2 + z2 + 2xy + 2yz + 2xz + 2x2 + xy + xz) - (y + z)( y2+ z2 - yz)
= (y+z)(x2 + y2 + z2 + 2xy + 2yz + 2xz + 2x2 + xy + xz - y2+ z2 - yz)
= (y+z)(3x2 + 3xy + 3yz + 3xz )
= 3(y+z)(x2 + xy + yz + xz )
= 3(y+z)[x(x+y) + z(x+y)]
= 3(x+y)(y+z)(x+z)
b) 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 +1+2009)
= (x2 +x +1)(x2 -x +2010)