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\(x^6-y^3=\left(x^2\right)^3-y^3=\left(x^2-y\right)\left(x^4+x^2y+y^2\right)\)
x6+3x4y2-8x3y3+3x2y4+y6= x6+3x4y2+3x2y4+y6-8x3y3=(x2+y2)3-(2xy)3
= (x2+y2-2xy)[(x2+y2)2+2xy(x2+y2)+(2xy)2]= (x-y)2(x4+6x2y2+y4+2x3y+2xy3)
(x2+y2-5)2-4x2y2-16xy-16=(x2+y2-5)2-(4x2y2+16xy+16)=(x2+y2-5)2-(2xy+4)2
=(x2+y2-5+2xy+4)(x2+y2-5-2xy-4)=(x2+2xy+y2-1)(x2-2xy+y2-9)=[(x+y)2-1][(x-y)2-32]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)
x4+324=x4+36x2+324-36x2=(x2+18)2-(6x)2=(x2+18-6x)(x2+18+6x)
a)
\(\left(y^3+8\right)+\left(y^2-4\right)\)
\(=\left(y+2\right)\left(y^2-2y+4\right)+\left(y-2\right)\left(y+2\right)\)
\(=\left(y+2\right)\left(y^2-2y+4+y+2\right)\)
\(=\left(y+2\right)\left(y^2+y+6\right)\)
b)
\(x^6-1=\)
\(=\left(x^3+1\right)\left(x^3-1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)\left(x-1\right)\left(x^2+x+1\right)\)
a. 2xy - x2 - y2 + 16
=(2xy-x2-y2)+16
=16-(x-y)2
=(4+x-y)(4-x+y)
b. x2 + x - 6
=x2+3x-2x-6
=x(x+3)-2(x+3)
=(x-2)(x+3)
c. x2 + 5x + 6
=x2+3x+2x+6
=x(x+3)+2(x+3)
=(x+2)(x+3)
Ta có : x6 - y6
\(=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
x6-y6=(x3)2-(y3)2=(x3-y3)(x3+y3)
=(x-y)(x2+xy+y2)(x+y)(x2-xy+y2)
đến đây bạn tự làm tiếp nha