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x4+2010x2+2009x+2010
=x4-x+2010x2+2010x+2010
=x.(x3-1)+2010.(x2+x+1)
=x.(x-1)(x2+x+1)+2010.(x2+x+1)
=(x2+x+1)(x2-x+2010)
(x+y+z)3-x3-y3-z3=(x+y+z-x)[(x+y+z)2+(x+y+z).x+x2]-(y+z)(y2-yz+z2)
=(y+z)(x2+y2+z2+2xy+2yz+2zx+x2+xy+zx+x2)-(y+z)(y2-yz+z2)
=(y+z)(3x2+y2+z2+3xy+2yz+3zx)-(y+z)(y2-yz+z2)
=(y+z)(3x2+y2+z2+3xy+2yz+3zx-y2+yz-z2)
=(y+z)(3x2+3yz+3xy+3zx)
=3.(y+z)(x2+xy+yz+zx)
=3.(y+z)[x.(x+y)+z.(x+y)
=3.(y+z)(x+y)(x+z)
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 = \(x^4+x^2+1+2009x^2+2009x+2009\)
\(=\left(x^4+2x^2+1\right)-x^2+2009\left(x^2+x+1\right)\)
\(=\left(x^2+1\right)^2-x^2+2009\left(x^2+x+1\right)\)
\(=\left(x^2+1-x\right)\left(x^2+1+x\right)+2009\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2010\right)\)
Đặt \(x+y-z=a;x-y+z=b;y+z-x=c\)
Ta có:\(A=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(A=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(A=\left(a+b\right)^3+3\left(a+b\right)\cdot c\cdot\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(A=a^3+b^3+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(A=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(A=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Hay \(A=3\cdot2x\cdot2y\cdot2z\)
\(A=24xyz\)
\(A=\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
Đặt \(x-y=a,y-z=b,z-x=c\Rightarrow a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+3ab.\left(-c\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Vậy \(A=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
\(x^4+4x^2+16\)
\(=\left(x^2\right)^2+2.x^2.4+4^2-4x^2\)
\(=\left(x^2+4\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+4\right)\left(x^2+2x+4\right)\)
a) x^3−3x^2−4x+12
=(x^3-3x^2)-(4x-12)
=x^2(x-3)-4(x-3)
=(x-3)(x^2-4)=(x-3)(x-2)(x+2)
b) x^4-5x^2+4=x^4-x^2-4x^2+4
=(x^4-x^2) - ( 4x^2-4)
=x^2(x^2-1) - 4(x^2-1)
=(x^2-1)(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
c) (x+y+z)^3-x^3-y^3-z^3
=x^3+y^3+z^3+3x^2yz+3xy^2z+3xyz^2-x^3-y^3-z^3
=3x^2yz+3xy^2z+3xyz^2
3xyz(x+y+z)
a. Câu hỏi của nguyễn khánh linh - Toán lớp 8 - Học toán với OnlineMath
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)