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x 8 + x 4 + 1 = x 8 + 2 x 4 + 1 – x 4 = ( x 8 + 2 x 4 + 1 ) – x 4 = [ ( x 4 ) 2 + 2 . x 4 . 1 + 12 ] – x 4 = ( x 4 + 1 ) 2 – ( x 2 ) 2 = ( x 4 + 1 – x 2 ) ( x 4 + 1 + x 2 ) = ( x 4 – x 2 + 1 ) ( x 4 + 2 x 2 – x 2 + 1 ) = ( x 4 – x 2 + 1 ) [ ( ( x 2 ) 2 + 2 . 1 . x 2 + 1 ) – x 2 ] = ( x 4 – x 2 + 1 ) [ ( x 2 + 1 ) 2 – x 2 ] = ( x 4 – x 2 + 1 ) ( x 2 + 1 – x ) ( x 2 + 1 + x ) = ( x 4 – x 2 + 1 ) ( x 2 – x + 1 ) ( x 2 + x + 1 )
Đáp án cần chọn là: C
a: \(x^4+4=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b: \(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
c: \(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4-x^2+1\right)\cdot\left(x^4+x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
Ta có
x 8 + 4 = x 4 2 + 4 x 4 + 4 - 4 x 4 = x 4 2 + 2 . 2 . x 2 + 2 2 - 2 x 2 2 = x 4 + 2 2 - 2 x 2 2
Đáp án cần chọn là: D
Lời giải:
a.
$3x^2+xy-4y^2=(3x^2-3xy)+(4xy-4y^2)=3x(x-y)+4y(x-y)=(x-y)(3x+4y)$
b.
$x^8-5x^4+4=(x^8-x^4)-(4x^4-4)$
$=x^4(x^4-1)-4(x^4-1)=(x^4-1)(x^4-4)$
$=(x^2-1)(x^2+1)(x^2-2)(x^2+2)$
$=(x-1)(x+1)(x^2+1)(x-\sqrt{2})(x+\sqrt{2})(x^2+2)$
c.
$x^3+3x^2+3x-7=(x^3+3x^2+3x+1)-8$
$=(x+1)^3-2^3=(x+1-2)[(x+1)^2+2(x+1)+4]$
$=(x-1)(x^2+4x+7)$
a) \(3x^2+xy-4y^2=3x^2-3xy+4xy-4y^2\)
\(=3x(x-y)+4y(x-y)=(3x+4y)(x-y)\)
b)\(x^8-5x^4+4=x^8-x^4-4x^4+4\)
\(=x^2(x^4-1)-4(x^4-1)=(x^2-4)(x^4-1)\)
\(=(x-2)(x+2)(x^2-1)(x^2+1)=(x-2)(x+2)(x-1)(x+1)(x^2+1)\)
c)\(x^3+3x^2+3x-7=x^3+3x^2+3x+1-8\)
\(\left(x+1\right)^3-\sqrt{2}^3=\left(x+1-\sqrt[]{2}\right)\left(\left(x+1\right)^2+2\sqrt{2}x+2\right)\)
Lời giải:
a.
$x^8+x^4+1=(x^4)^2+2x^4+1-x^4$
$=(x^4+1)^2-(x^2)^2=(x^4+1-x^2)(x^4+1+x^2)$
$=(x^4+1-x^2)[(x^2+1)^2-x^2]$
$=(x^4-x^2+1)(x^2+1-x)(x^2+1+x)$
b.
$x^{12}-3x^6-1=(x^6-\frac{3}{2})^2-\frac{13}{4}$
$=(x^6-\frac{3}{2}-\frac{\sqrt{13}}{2})(x^6-\frac{3}{2}+\frac{\sqrt{13}}{2})$
c.
$3x^4+10x^2-25=(3x^4+15x^2)-(5x^2+25)$
$=3x^2(x^2+5)-5(x^2+5)=(x^2+5)(3x^2-5)$
$=(x^2+5)(\sqrt{3}x-\sqrt{5})(\sqrt{3}x+\sqrt{5})$
c.
$x^2-5y^2-y^4+2xy-9$
$=(x^2+2xy+y^2)-(y^4+6y^2+9)$
$=(x+y)^2-(y^2+3)^2$
$=(x+y+y^2+3)(x+y-y^2-3)$
\(a,x^8+x^4+1\\ =\left(x^8+2x^4+1\right)-x^4\\ =\left(x^4+1\right)^2-x^4\\ =\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\\ b,x^{12}-3x^6-1\\ =\left(x^{12}-2x^6+1\right)-x^6-2\\ =\left(x^6-1\right)^2-x^6-2\\ =\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)-2???\\ c,3x^4+10x^2-25\\ =4x^4-\left(x^4-10x^2+25\right)\\ =4x^4-\left(x^2-5\right)^2\\ =\left(2x^2-x^2+5\right)\left(2x^2+x^2-5\right)\\ =\left(x^2+5\right)\left(3x^2-5\right)\\ d,x^2-5y^2-y^4+2xy-9\\ =\left(x^2+2xy+y^2\right)-\left(y^4+6y^2+9\right)\\ =\left(x+y\right)^2-\left(y^2+3\right)^2\\ =\left(x+y+y^2+3\right)\left(x+y-y^2-3\right)\)
a. $6x^2-11x=x(6x-11)$
b. $x^7+x^5+1=(x^7-x)+(x^5-x^2)+x+x^2+1$
$=x(x^6-1)+x^2(x^3-1)+(x^2+x+1)$
$=x(x^3-1)(x^3+1)+x^2(x^3-1)+(x^2+x+1)$
$=(x^3-1)(x^4+x+x^2)+(x^2+x+1)$
$=(x-1)(x^2+x+1)(x^4+x^2+x)+(x^2+x+1)$
$=(x^2+x+1)[(x-1)(x^4+x^2+x)+1]$
$=(x^2+x+1)(x^5-x^4+x^3-x+1)$
c.
$x^8+x^4+1=(x^4)^2+2.x^4+1-x^4$
$=(x^4+1)^2-(x^2)^2$
$=(x^4+1-x^2)(x^4+1+x^2)$
$=(x^4+1-x^2)(x^4+2x^2+1-x^2)$
$=(x^4-x^2+1)[(x^2+1)^2-x^2]$
$=(x^4-x^2+1)(x^2+1-x)(x^2+1+x)$
d.
$x^3-5x+8-4=x^3-5x+4$
$=x^3-x^2+x^2-x-(4x-4)$
$=x^2(x-1)+x(x-1)-4(x-1)=(x-1)(x^2+x-4)$
e.
$x^5+x^4+1=(x^5-x^2)+(x^4-x)+x^2+x+1$
$=x^2(x^3-1)+x(x^3-1)+x^2+x+1$
$=(x^3-1)(x^2+x)+(x^2+x+1)$
$=(x-1)(x^2+x+1)(x^2+x)+(x^2+x+1)$
$=(x^2+x+1)[(x-1)(x^2+x)+1]$
$=(x^2+x+1)(x^3-x+1)$
x8 + 98x4 + 1
= (x8 + 2x4 + 1 ) + 96x4
= (x4 + 1)2 + 16x2(x4 + 1) + 64x4 - 16x2(x4 + 1) + 32x4
= (x4 + 1 + 8x2)2 – 16x2(x4 + 1 – 2x2)
= (x4 + 8x2 + 1)2 - 16x2(x2 – 1)2
= (x4 + 8x2 + 1)2 - (4x3 – 4x )2
= (x4 + 4x3 + 8x2 – 4x + 1)(x4 - 4x3 + 8x2 + 4x + 1)
x8 + 98x4 + 1
= (x8 + 2x4 + 1 ) + 96x2
= (x4 + 1)2 + 16x2(x4 + 1) + 64x4 - 16x2(x4 + 1) + 32x4
= (x4 + 1 + 8x2)2 – 16x2(x4 + 1 – 2x2)
= (x4 + 8x2 + 1)2 - 16x2(x2 – 1)2
= (x4 + 8x2 + 1)2 - (4x3 – 4x )2
= (x4 + 4x3 + 8x2 – 4x + 1)(x4 - 4x3 + 8x2 + 4x + 1)
Vậy...