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a: =64x^4+16x^2y^2+y^4-16x^2y^2

=(8x^2+y^2)^2-(4xy)^2

=(8x^2+y^2-4xy)(8x^2+y^2+4xy)

b: =x^8+2x^4+1-x^4

=(x^4+1)^2-x^4

=(x^4-x^2+1)(x^4+x^2+1)

=(x^4-x^2+1)(x^4+2x^2+1-x^2)

=(x^4-x^2+1)(x^2+1-x)(x^2+x+1)

c: =(x+1)(x^2-x+1)+2x(x+1)

=(x+1)(x^2-x+1+2x)

=(x+1)(x^2+x+1)

d: =(x^2-1)(x^2+1)-2x(x^2-1)

=(x^2-1)(x^2-2x+1)

=(x-1)^2*(x-1)(x+1)

=(x+1)(x-1)^3

15 tháng 7 2023

\(a)x^5+x^4+1\)

\(=x^3\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^3-x+1\right)\)

\(b)x^8+x^7+1\)

\(=\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

\(#Tuyết\)

17 tháng 7 2021

a) x12 + 4 = x12 + 4x6 + 4 - 4x6 = (x6 + 2)2 - (2x3)2 

= (x6 - 2x3 + 2)(x6 + 2x3 + 2)

b) 4x8 + 1 = 4x8 + 4x4  + 1 - 4x4 = (2x4 + 1)2 - (2x2)2 

= (2x4 + 2x2 + 1)(2x4 - 2x2  + 1)

17 tháng 7 2021

c) x7 + x5 - 1 = x7 - x + x5 + x2 - (x2 - x  + 1) = x(x6 - 1) + x2(x3 + 1) - (x2 - x + 1)

= x(x3 - 1)(x3 + 1) + x2(x + 1)(x2 - x + 1) - (x2 - x + 1)

= (x4 - x)(x + 1)(x2 - x + 1) + (x3 + x2)(x2 - x + 1) - (x2 - x + 1)

= (x5 + x4 - x2 - x + x3 + x2 - 1)(x2 -x + 1)

= (x5 + x4 + x3 - x - 1)(x2 - x + 1)

d) x+ x5 + 1 = x7 - x + x5 - x2 + (x2 + x + 1)

= x(x3 - 1)((x3 + 1) + x2(x3 - 1) + (x2 + x + 1)

= (x4 + x)(x  - 1)(x2 + x + 1) + x2(x - 1)((x+ x + 1) + (x2 + x + 1)

= (x2 + x + 1)(x5 - x4 + x- x + x3 - x2 + 1)

= (x2 + x + 1)(x5 - x4 + x3 - x + 1)

17 tháng 8 2020

a) \(x^5-x^4-1\)

\(=\left(x^5+x^2\right)-\left(x^4+x\right)-\left(x^2-x+1\right)\)

\(=x^2\left(x^3+1\right)-x\left(x^3+1\right)-\left(x^2-x+1\right)\)

\(=x^2\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\)

\(=\left(x^2-x+1\right)\left(x^3+x^2-x^2-x-1\right)\)

\(=\left(x^2-x+1\right)\left(x^3-x-1\right)\)

17 tháng 8 2020

b) \(x^8+x^7+1\)

\(=\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\)

\(=x^2\left(x^6-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)\)

\(=x^2\left(x^3-1\right)\left(x^3+1\right)+x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)

\(=x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[\left(x^3-x^2\right)\left(x^3+1\right)+\left(x^2-x\right)\left(x^3+1\right)+1\right]\)

\(=\left(x^2+x+1\right)\left[\left(x^3-x\right)\left(x^3+1\right)+1\right]\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

a) Ta có: \(x^4-16x^2=0\)

\(\Leftrightarrow x^2\left(x^2-16\right)=0\)

\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

b) Ta có: \(x^8+36x^4=0\)

\(\Leftrightarrow x^4\left(x^4+36\right)=0\)

\(\Leftrightarrow x^4=0\)

hay x=0

c) Ta có: \(\left(x-5\right)^3-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left[\left(x-5\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

d) Ta có: \(5\left(x-2\right)-x^2+4=0\)

\(\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

22 tháng 11 2017

x8+x7+1= x8+x7+x6-x6-x5-x4+x5+x4+x3-x3-x2-x+x2+x+1

=(x8+x7+x6)-(x6+x5+x4)+(x5+x4+x3)-(x3+x2+x)+(x2+x+1)

= x6(x2+x+1)-x4(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+( x2+x+1)

=(x2+x+1)(x6-x4+x3-x+1)

Câu b, c lm tương tự

5 tháng 7 2015

(x+2)(x+3)(x+4)(x+5) - 8

=(x+2)(x+5)(x+3)(x+4)-8

=(x2+7x+10)(x2+7x+12)-8

đặt t=x2+7x+10 ta được:

t(t+2)-8=t2+2t-8

=t2-2t+4t-8

=t(t-2)+4(t-2)

=(t-2)(t+4)

thay t=x2+7x+10 ta được:

(x2+7x+8)(x2+7x+14)

vậy  (x+2)(x+3)(x+4)(x+5) - 8=(x2+7x+8)(x2+7x+14)

29 tháng 10 2018

\(x^8+x^7+1\)

\(=\left(x^8-x^6+x^5-x^3+x^2\right)+\left(x^7-x^5+x^4-x^2+x\right)+\left(x^6-x^4+x^3-x+1\right)\)

\(=x^2\left(x^6-x^4+x^3-x+1\right)+x\left(x^6-x^4+x^3-x+1\right)+\left(x^6-x^4+x^3-x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

2 tháng 11 2018

\(x^5-x^4-1\)

\(=x^5-x^3-x^2-x^4+x^2+x+x^3-x-1\)

\(=x^2\left(x^3-x-1\right)-x\left(x^3-x-1\right)+\left(x^3-x-1\right)\)

\(=\left(x^2-x+1\right)\left(x^3-x-1\right)\)

19 tháng 11 2016

a, x8 + x7 + 1

=x2 (x6 - 1) + x (x6 - 1) +(x2 + x + 1)

= (x6 _ 1)(x2 + x) + (x2 + x +1)

= (x3 - 1)(x3 + 1)( x2 + x) + (x2 + x +1)

=(x - 1)(x2 + x +1)( x2 + x) + (x2 + x +1)

=(x2 + x +1)((x - 1)( x2 + x) +1)

=(x2 + x +1)(x3 + 1)

b, x5 - x4-1

c, x7+x5 + 1

d,x8 + x4 +1

Chú ý: Các đa thức có dạng: x3m+1+x3n+2+1 như x7+x2+1; x7+x5+1; x8 + x4 +1;

x5+x+1; x8+x+1 đều có nhân tử chung là x2 + x +1

Các phần còn lại tương tự nhé!!!

19 tháng 11 2016

cảm ơn ạ