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x8 + x4 + 1
=x8+2x4+1-x4
=(x4+1)2-x4
=(x4-x2+1)(x4+x2+1)
=(x4-x2+1)(x4+2x2+1-x2)
=(x4-x2+1)[(x2+1)2-x2]
=(x4-x2+1)(x2-x+1)(x2+x+1)
x8 + x4 + 1
= ( x4 )2 + 2x4 + 1 - x4
= ( x4 + 1 )2 - x4
= ( x4 + 1 - x2 ) ( x4 + 1 + x2 )
1) \(x^3+x^2+4\)
\(=\left(x^3-x^2+2x\right)+\left(2x^2-2x+4\right)\)
\(=x\left(x^2-x+2\right)+2\left(x^2-x+2\right)\)
\(=\left(x^2-x+2\right)\left(x+2\right)\)
2) \(x^3-2x-4\)
\(=\left(x^3+2x^2+2x\right)-\left(2x^2+4x+4\right)\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x^2+2x+2\right)\left(x-2\right)\)
x4+2011x2+2010x+2011
=(x4+x3+x2)+(2011x2+2011x+2011)-(x3+x2+x)
=x2(x2+x+1)+2011(x2+x+1)-x(x2+x+1)
=(x2+x+1)(x2+2011-x)
x4+2011x2+2010x+2011=x4-x+2011x2+2011x+2011
=x(x3-1)+2011(x2+x+1)
=x(x- 1)(x2+x+1)+2011(x2+x+1)
=(x2+x+1)[x(x-1)+2011]
=(x2+x+1)(x2-x+2011)
=[(x+2)(x+5)][(x+3)(x+4)]−24=(x2+7x+10)(x2+7x+12)−24=(x2+7x+11)2−1−24=(x2+7x+11)2−25=(x2+7x+11−5)(x2+7x+11+5)=(x2+7x+6)(x2+7x+16)=(x+1)(x+6)(x2+7x+16)
\(2x^2y^3-\frac{x}{4}-4y^6\)
đây là pt bậc 2 của y^3 , ta đặt y^3=z ta được
\(-\left(4z^2+\frac{2.2xz}{2}+\frac{x^2}{4}\right)+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)
\(-\left(2z+\frac{x}{2}\right)^2+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)
\(-\left\{\left(2x+\frac{x}{2}\right)^2-\left(\frac{x^2}{4}-\frac{x}{4}\right)\right\}\)
\(-\left\{\left(2x+\frac{x}{2}+\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\left(2x+\frac{x}{2}-\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\right\}\)
Có:\(x^4+64y^4\)
\(=\left(x^4+16x^2y^2+64y^4\right)-16x^2y^2\)
\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)
\(=\left(x^2+4xy+8y^2\right)\left(x^2-4xy+8y^2\right)\)
Linz
= 64y4 + 32xy3 + 8y2x2 - 32xy3 -16x2y2 - 4x3y + 8x2y2 +4x3y +x4
= 8y2 ( 8y2 + 4xy + x2 ) - 4xy ( 8y2 + 4xy + x2 ) + x2 ( 8y2 + 4xy + x2 )
= ( 8y2 - 4xy + x2 ) ( 8y2 + 4xy + x2 )