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\(x^5+x^4+2\)
\(=x^5+x^4+x^2-x^2+1+1\)
\(=\left(x^5-x^2\right)+\left(x^4+x^2+1\right)\)
\(=\left(x^5-x^2\right)+\left(x^4+2x^2-x^2+1\right)+1\)
\(=x^2\left(x^3-1\right)+\left(x^4+2x^2-x^2+1\right)+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(\left(x^2+1\right)^2-x^2\right)+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+1+x\right)\cdot\left(x^2+1-x\right)+1\)
\(=\left(x^3-x^2\right)\left(x^2+x+1\right)+\left(x^2+1+x\right)\cdot\left(x^2+1-x\right)+1\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+x^2+1-x\right)+1\)
\(=\left(x^2+x+1\right)\left(x^3+1-x\right)+1\)
<=>x4-x+x2 +x+1= x (x-1) (x2+x+1) + (x2+x+1) = (x2+x+1)(x2-x+1)
chắc có lẽ đúng đó
Nghịch xíu :v
a, \(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)-2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x+2\right)\)
b, \(x^2+4x+3\)
\(=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
Chúc bạn học tốt!!!
\(x^4+3x^2-4\)
\(=x^4+4x^2-x^2-4\)
\(=x^2\left(x^2+4\right)-\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x^2-1\right)\)
\(=\left(x^2+4\right)\left(x-1\right)\left(x+1\right)\)
Chúc bạn học tốt.
Ta có: x3 - x2 - 4
= (x3 - 1) - (x2 - 2x + 1) - 2(x + 1)
= (x - 1)(x2 + x + 1) - (x - 1)2 - 2(x + 1)
= (x - 1)(x2 + x + 1 - x + 1 - 2)
= x2(x - 1)
x3 - x2 - 4
= x3 + x2 - 2x2 - 4
= (x3 - 2x2) + (x2 - 4)
= x2 (x - 2) - (x2 - 22)
= x2 (x - 2) - (x + 2) (x - 2)
= (x - 2) [x2 + (x + 2)]
= (x - 2) (x2 + x + 2)
#Học tốt!!!
~NTTH~
\(x^4+x^2+1\)
\(=x^4+2x^2+1+x^2-2x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2+1-x\right).\left(x^2+1+x\right)\)
Vì phương trình x4+x2+1=0 vô nghiệm nên không thể phân tích thành nhân tử
Ta có : x4 + x2 + 1
= x4 + x2 + x2 + 1 - x2
= (x2 + 1)2 - x2
= (x2 + 1 - x)(x2 + 1 + x)
x4 + x2 + 1
= x4 + 2x2 + 1 - x2
= ( x2 + 1 )2 - x2
= ( x2 - x + 1 )( x2 + x + 1 )
\(x^2-2x-4=x^2-2x+1-5\)
\(=\left(x-1\right)^2-5=\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right)\)
Trl :
\(x^2-2x-4=x^2-2x+1-5\)
\(\Rightarrow\left(x-1\right)^2-5=\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right)\)
Mình xin lỗi nhé, để mình sửa lại : ^^
a) \(x^4+3x^2+4=\left(x^4+x^3+2x^2\right)+-\left(x^3+x^2+2x\right)+2\left(x^2+2x+2\right)\)
\(=x^2\left(x^2+x+2\right)-x\left(x^2+x+2\right)+2\left(x^2+x+2\right)=\left(x^2-x+2\right)\left(x^2+x+2\right)\)
b) \(x^4+5x^2+9=\left(x^4+x^3+3x^2\right)-\left(x^3+x^2+3x\right)+3\left(x^2+x+3\right)\)
\(=x^2\left(x^2+x+3\right)-x\left(x^2+x+3\right)+3\left(x^2+x+3\right)=\left(x^2-x+3\right)\left(x^2+x+3\right)\)
x2+4
=x2+4+4x-4x
=(x2+2.x.2+22)-4x
=(x+2)2-(2√x)2
=(x+2-2√x)(x+2+2√x)