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= x^10 - x + x^5 - x^2 + x^2 + x + 1
= x ( x^9 - 1 ) + x^2 (x^3 - 1 ) + x^2 + x + 1
= x [ ( x^3 - 1) ( x^6 + x^3 + 1 )] + x^2 ( x - 1 )(x^2 + x + 1 ) + x^2 + x + 1
= x ( x - 1 )(x^2 + x + 1 )(x^6 + x^3 + 1) + x^2 (x-1 )(x^2 + x+ 1 ) + x^2 + x + 1
= (x^2 + x + 1 )[ x(x-1)(x^6 + x^3 + 1 ) + x^2 + 1 )
Nhân ra giúp mình nha
x10 + x5 + 1 = (x10 - x) + (x5 - x2) + (x2 + x + 1) = x.[(x3)3 - 1] + x2.(x3 - 1) + (x2 + x + 1)
= x.(x3 - 1).(x6 + x3 + 1) + x2.(x3 - 1) + (x2 + x + 1)
= (x2 + x + 1). [x.(x -1).(x6 + x3 + 1) + x2 + 1 ]
\(x^5+x+1=x^5-x^2+x^2+x+1=x^2\left(x^3-1\right)+\left(x^2+x+1\right)=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
\(x^{10}+x^5+1=x^{10}-x+x^5-x^2+x^2+x+1=x\left(x^9-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3-1\right)\left(x^6+x^3+1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)\left(x^6+x^3+1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)\left(x^6+x^3+1\right)+x^2+1\right]\)
\(=\left(x^2+x+1\right)\left(x^8-x^7+x^5-x^4+x^3-x+1\right)\)
\(A\) \(=\) \(x^{10}+x^5+1\)
\(A=\left(x^{10}+x\right)+\left(x^5-^2\right)+\left(x^2+x+1\right)\)
\(A=x\left(x^3-1\right)\left(x^6+x^3+1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(A=\left(x^2+x+1\right)\left(x^8-x^7+x^5-x^4+x^3-x+1\right)\)
hơi tắt các bạn tự hiểu nhé
(thanks)
a^10-a^7+a^7-a^4+a^4-a+a+a^5-a^2+a^2+1
=(a^10-a^7)+(a^7-a^4)+(a^4-a) + (a^5-a^2) + (a^2+a+1)
=a^7(a^3-1)+a^4(a^3-1) +a(a^3-1)+a^2(a^3-1) + (a^2+a+1)
=(a^7+a^4+a+a^2)(x-1)(a^2+a+1)+(a^2+a+1)
Bạn làm tiếp đặt a^2+a+1 làm nhân tử chung ..các câu sau cũng như thế nhé ^.^
Làm được câu a thôi nhé
Cách 1:
a10 + a5 + 1
= a10 - a9 + a7 - a6 + a5 - a3 + a2 + a9 - a8 + a6 - a5 + a4 - a3 + a + a8 - a7 + a5 - a4 + a2 - a + 1
nhóm 7 hạng tử ta đc :
= a2(a8 - a7 + a5 - a4 + a3 - a + 1) + a(a8 - a7 + a5 - a4 + a3 - a + 1) + (a8 - a7 + a5 - a4 + a3 - a + 1)
= (a2 + a + 1)(a8 - a7 + a5 - a4 + a3 - a + 1)
Cách 2:
x10 + x5 + 1 = (x10 - x) + (x5 - x2) + (x2 + x + 1)
= x.[(x3)3 - 1] + x2.(x3 - 1) + (x2 + x + 1)
= x.(x3 - 1).(x6 + x3 + 1) + x2.(x3 - 1) + (x2 + x + 1)
= (x2 + x + 1). [x.(x -1).(x6 + x3 + 1) + x2 + 1 ]
P.s:Ko chắc ^^!
a)\(x^2+7x+12\)
\(=x^2+x+6x+6\)
\(=x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x+6\right)\)
a) \(x^5+x-1\)
\(=x^5+x^4+x^3+x^2-x^4-x^3-x^2+x-1\)
\(=\left(x^5-x^4+x^3\right)+\left(x^4-x^3+x^2\right)-\left(x^2-x+1\right)\)
\(=x^3\left(x^2-x+1\right)+x^2\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3+x^2-1\right)\)(còn 1 cách nữa là thêm bớt \(x^2\)vào bạn nhé!)
b) \(x^7+x^2+1\)
\(=x^7-x+x^2+x+1\)
\(=x\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3+1\right)\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x^3+1\right)\left(x-1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
(Chúc bạn học tốt và nhớ tíck cho mình với nhé!)
\(x^{10}+x^2+1\)
\(=x^{10}+x^8-x^8+x^6-x^6+x^4-x^4+x^2+1\)
\(=\left(x^{10}+x^8+x^6\right)-\left(x^8+x^6+x^4\right)+\left(x^4+x^2+1\right)\)
\(=x^6\left(x^4+x^2+1\right)-x^4\left(x^4+x^2+1\right)+\left(x^4+x^2+1\right)\)
\(=\left(x^6-x^4+1\right)\left(x^4+x^2+1\right)\)
\(=\left(x^6-x^4+1\right)\left(x^4+x^3-x^3+x^2+x^2-x^2+x-x+1\right)\)
\(=\left(x^6-x^4+1\right)\)
\(\left[\left(x^4-x^3+x^2\right)+\left(x^3-x^2+x\right)+\left(x^2-x+1\right)\right]\)
\(=\left(x^6-x^4+1\right)\)
\(\left[x^2\left(x^2-x+1\right)+x\left(x^2-x+1\right)+\left(x^2-x+1\right)\right]\)
\(=\left(x^6-x^4+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)
x10+x2+1
=( x10 - x ) + ( x2 + x + 1)
= x[ (x3)3-1] + ( x2 + x +1)
=x[( x3-1)( x6 + x3 +1) + (x2 + x +1)
=x[(x-1)(x2 + x +1)( x6 + x3 +1)] + (x2 + x +1)
=x(x2 + x +1)[(x-1)( x6 + x3 +1) +1 ]
=x2(x2 + x +1)(x6-x5+x3-x2+1)
a, x10+x9+x8-x9-x8-x7+x7+x6+x5-x6-x5-x4+x5+x4+x3-x3-x2-x+x2+x+1 = x8(x2+x+1)-x7(x2+x+1)+x5(x2+x+1)-x4(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+(x2+x+1) =(x8-x7+x5-x4+x3-x+1)
b,x8+x7+x6-x7-x6-x5+x5+x4+x3-x3-x2-x+x2+x+1 =x6( x2+x+1)-x5(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+(x2+x+1) = (x2+x+1)(x6-x5+x3-x+1)
a)Ta có: x10+x5+1=x10+x7-x7+x6-x6+x5+1
=(x10-x7) - (x6-1) + (x7+x6+x5)
=x7(x3-1) - ((x3)2-1) + (x2+x+1)
=x7(x-1)(x2+x+1) - (x3-1)(x3+1) + x5(x2+x+1)
=x7(x-1)(x2+x+1) - (x-1)(x2+x+1)(x3+1) + x5(x2+x+1)
=(x2+x+1)(x7(x+1)-(x+1)(x3+1)+x5)
=(x2+x+1)(x8-x7+x5-x4+x3-x+1)