PHÂN TÍCH ĐA THỨC THÀNH NHÂN TỬ
X5 + X4 + 1
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x⁴ - 2x³ + 2x - 1
= (x⁴ - 1) - (2x³ - 2x)
= (x² - 1)(x² + 1) - 2x(x² - 1)
= (x² - 1)(x² + 1 - 2x)
= (x - 1)(x + 1)(x² - 2x + 1)
= (x - 1)(x + 1)(x - 1)²
= (x - 1)³(x + 1)
Lời giải:
a.
$=(x^2)^2+(\frac{1}{2}y^4)^2+2.x^2.\frac{1}{2}y^4-x^2y^4$
$=(x^2+\frac{1}{2}y^4)^2-(xy^2)^2$
$=(x^2+\frac{1}{2}y^4-xy^2)(x^2+\frac{1}{2}y^4+xy^2)$
b.
$=(\frac{1}{2}x^2)^2+(y^4)^2+2.\frac{1}{2}x^2.y^4-x^2y^4$
$=(\frac{1}{2}x^2+y^4)^2-(xy^2)^2$
$=(\frac{1}{2}x^2+y^4-xy^2)(\frac{1}{2}x^2+y^4+xy^2)$
c.
$=(8x^2)^2+(y^2)^2+2.8x^2.y^2-16x^2y^2$
$=(8x^2+y^2)^2-(4xy)^2=(8x^2+y^2-4xy)(8x^2+y^2+4xy)$
d.
$=\frac{64x^4+y^4}{64}=\frac{1}{64}(8x^2+y^2-4xy)(8x^2+y^2+4xy)$
c: \(64x^4+y^4\)
\(=64x^4+16x^2y^2+y^4-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)
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^3(x-y)+(x-y)
=(x-y)(x^3+1)
=(x-y)(x+1)(x^2-x+1)
b: =(a-1)^2-9b^2
=(a-1-3b)(a-1+3b)
x4+4 = (x2)2+22 = x4 + 2.x2.2 + 4 – 4x2
= (x2 + 2)2 – (2x)2 = (x2-2x+2)(x2+2x+2)
Ta có: \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)
\(A=x^4+4\)
\(=\) \(x^4+4+4x^2-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(A=\) \(\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
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: \(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4-x^2+1\right)\cdot\left(x^4+x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
a/ 2x^2 (x – 1) + 4x (1 – x)
= 2x^2(x – 1) – 4x (x – 1)
= (x – 1)( 2x^2 – 4x)
=2x(x – 1)(x – 2)
\(x^5+x^4+1\)
\(=x^5-x^3+x^3+x^4+1\)
\(=x^5+x^4+x^3-\left(x^3-1\right)\)
\(=x^3\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^3-x+1\right)\left(x^2+x+1\right)\)
bạn ơi cho mình hỏi tí
tại sao :
\(\left(x^{^3}-1\right)=\left(x-1\right)\left(x^{^2}-x+1\right)\)
mình k cho bn rồi đó