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x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
x5-x4-1=x5-x3-x2-x4+x2+x+x3-x-1
=x2.(x3-x-1)-x.(x3-x-1)+(x3-x-1)
=(x3-x-1)(x2-x+1)
x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
(x+1).(x+2).(x+3).(x+4)-4
=(x+1)(x+4)(x+2)(x+3)-4
=(x2+5x+4)(x2+5x+6)-4
Đặt t=x2+5x+4 ta được:
t.(t+2)-4
=t2+2t-4
Vẫn sai đề
#)Giải :
\(\left(x+1\right)^4+\left(x^2+x+1\right)^2\)
\(=\left(x+1\right)^4+x^2\left(x+1\right)^2+2x\left(x+1\right)+1\)
\(=\left(x+1\right)^2\left(2x^2+2x+1\right)+\left(2x^2+2x+1\right)\)
\(=\left(2x^2+2x+1\right)\left(x^2+2x+2\right)\)
\(\left(x+1\right)^4+\left(x^2+x+1\right)^2\)
\(=[\left(x+1\right)^2-x^2-x-1]\left[\left(x+1\right)^2+x^2+x+1\right]\)
\(=(x^2+2x+1-x^2-x-1)(x^2+x+1+x^2+x+1)\)
\(=x\left(2x^2+2x+2\right)\)
\(=2x\left(x^2+x+1\right)\)
\(x^2\left(1-x^2\right)-4-4x^2\)
\(=-x^4+x^2-4-4x^2\)
\(=-\left(x^4+4+4x^2-x^2\right)\)
\(=-\left(\left(x^2+2\right)^2-x^2\right)\)
\(=-\left(x^2+2-x\right)\left(x^2+2+x\right)\)
\(x^2\left(1-x^2\right)-4-4x^2=x^2\left(1-x\right)\left(1+x\right)-4\left(1+x^2\right)\)
Đến đấy tách thế nào đây ( đề sai hả )
=[(x+1)(x+4)][(x+2)(x+3)]+8=(x2+5x+4)(x2+5x+6)+8
Đặt x2+5x+4=t
Ta có : t(t+2)+8=t2+2t-8=(t-2)(t+4)
k mk nha
#)Giải :
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+4x^3+x^2\right)+2\left(2x^2+x\right)+1\)
\(=\left(2x^2+x\right)^2+2\left(2x^2+x\right)+1\)
\(=\left(2x^2+x+1\right)^2\)
\(4x^4+4x^3+5x^2+2x+1=\left(4x^4+4x^3+x^2\right)+2\left(2x^2+x\right)+1\)
\(=\left(2x^2+x\right)^2+2\left(2x^2+x\right)+1\)
\(=\left(2x^2+x+1\right)^2\)
~ Rất vui vì giúp đc bn ~
Ta phân tích như sau e nhé :)
\(3x^4+3x^2+3-\left(x^4+x^2+2x+1+2x^2\left(x+1\right)\right)\)
\(=3x^4+3x^2+3-\left(x^4+x^2+2x+1+2x^3+2x^2\right)\)
\(=2x^4-2x^3-2x+2=2\left[x^3\left(x-1\right)-\left(x-1\right)\right]=2\left(x-1\right)^2\left(x^2+x+1\right)\)
x^4-x^2-1=(x^4-x)-(x^2+x+1)
=x(x^3-1)-(x^2+x+1)
=x(x-1)(x^2+x+1)-(x^2+x+1)
=(x^2+x+1)\([\)x(x-1)-1\(]\)
=(x^2+x+1)(x^2-x-1)