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AH
Akai Haruma
Giáo viên
7 tháng 9 2021
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.
$x^2-5y^2-y^4+2xy-9$
$=(x^2+2xy+y^2)-(y^4+6y^2+9)$
$=(x+y)^2-(y^2+3)^2$
$=(x+y+y^2+3)(x+y-y^2-3)$
7 tháng 9 2021
\(a,x^8+x^4+1\\ =\left(x^8+2x^4+1\right)-x^4\\ =\left(x^4+1\right)^2-x^4\\ =\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\\ b,x^{12}-3x^6-1\\ =\left(x^{12}-2x^6+1\right)-x^6-2\\ =\left(x^6-1\right)^2-x^6-2\\ =\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)-2???\\ c,3x^4+10x^2-25\\ =4x^4-\left(x^4-10x^2+25\right)\\ =4x^4-\left(x^2-5\right)^2\\ =\left(2x^2-x^2+5\right)\left(2x^2+x^2-5\right)\\ =\left(x^2+5\right)\left(3x^2-5\right)\\ d,x^2-5y^2-y^4+2xy-9\\ =\left(x^2+2xy+y^2\right)-\left(y^4+6y^2+9\right)\\ =\left(x+y\right)^2-\left(y^2+3\right)^2\\ =\left(x+y+y^2+3\right)\left(x+y-y^2-3\right)\)
26 tháng 8 2021
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: \(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4-x^2+1\right)\cdot\left(x^4+x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
TL
6
28 tháng 6 2021
11) Ta có: \(a^6+a^4+a^2b^2+b^4-b^6\)
\(=a^6-b^6+a^4+a^2b^2+b^4\)
\(=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)+\left(a^4+a^2b^2+b^4\right)\)
\(=\left(a^4+a^2b^2+b^4\right)\left(a^2-b^2+1\right)\)
12) Ta có: \(x^3+3xy+y^3-1\)
\(=\left(x^3+3x^2y+3xy^2+y^3-1\right)-3x^2y-3xy^2+3xy\)
\(=\left[\left(x+y\right)^3-1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[x^2+2xy+y^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)
14) Ta có: \(x^8+x+1\)
\(=x^8+x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4+x^3-x^3+x^2-x^2+x+1\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
15) Ta có: \(x^8+3x^4+4\)
\(=x^8+4x^4+4-x^4\)
\(=\left(x^4+2\right)^2-\left(x^2\right)^2\)
\(=\left(x^4-x^2+2\right)\left(x^4+x^2+2\right)\)
CM
16 tháng 7 2018
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) \(x^8+14x^4+1=\left(x^4+1\right)^2+12x^4=\left(x^4+1\right)^2+4x^2\left(x^4+1\right)+4x^4-4x^2\left(x^4+1\right)+8x^4\)
\(=\left(x^4+2x^2+1\right)^2-4x^2\left(x^4-2x^2+1\right)=\left(x^4+2x^2+1\right)^2-\left(2x\left(x^2-1\right)\right)^2\)
\(=\left(x^4-2x^3+2x^2+2x+1\right)\left(x^4+2x^3+2x^2-2x+1\right)\)