Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
<=>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
x^4-x^3-x^2+2x-2
=(x^4-x^3)-(x^2-2x+2)
=x^3(x-1)-(x-1)^2
=(x^3-x-1)*(x-1)
\(x^2\left(x^2+4\right)-x^2+4\)
\(=x^4+4x^2-x^2+4\)
\(=x^4+3x^2+4\)
\(=x^4-x^3+x^3+2x^2+2x^2-x^2-2x+2x+4\)
\(=\left(x^4-x^3+2x^2\right)+\left(x^3-x^2+2x\right)+\left(2x^2-2x+4\right)\)
\(=x^2\left(x^2-x+2\right)+x\left(x^2-x+2\right)+2\left(x^2-x+2\right)\)
\(=\left(x^2+x+2\right)\left(x^2-x+2\right)\)
\(x^2\left(x^2+4\right)-x^2+4=x^4+4x^2-x^2+4=x^4+3x^2+4\)
\(=\left(x^4+4x^2+4\right)-x^2\)
\(=\left(x^2+2\right)^2-x^2\)
\(=\left(x^2+x+2\right)\left(x^2-x+2\right)\)
Ta có : \(F=x^2-4^x+4-y^2\)
\(=\left(x^2-4^x+4\right)-y^2\)( nhóm hạng tử )
\(=\left(x-2\right)^2-y^2\)( đẳng thức số 2 )
\(=\left(x-2-y\right)\left(x-2+y\right)\)( đẳng thức số 3 )
Vậy : \(F=\left(x-2-y\right)\left(x-2+y\right)\)
\(x^5+x^4-x^3+x^2-x+2\)
\(=x^5+2x^4-x^4+2x^3-x^3+2x^2-x^2+2x-x+2\)
\(=x^4\left(x-2\right)-x^3\left(x-2\right)-x^2\left(x-2\right)-x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(x^4-x^3-x^2-x-1\right)\)
\(x^4+x^3+x^2-1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\left(x-1\right)\)
\(=\left(x+1\right)\left(x^3+\left(x-1\right)\right)\)
Ủng hộ nha ^ _ ^
\(x^4+x^3+x^2-1\)
\(=x^2\left(x^2-1\right)+x^2-1\)
\(=\left(x^2+1\right)\left(x^2-1\right)\)
\(x^5+x^4-x^3+x^2-x+2\)
\(=x^5-x^4+x^3-x^2+x+2x^4-2x^3+2x^2-2x+2\)
\(=x\left(x^4-x^3+x^2-x+1\right)+2\left(x^4-x^3+x^2-x+1\right)\)
\(=\left(x+2\right)\left(x^4-x^3+x^2-x+1\right)\)
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~
x2−4+x(x−2)=x2−4+x2−2x
x2−4+x2−2x=2x2−2x−4
2x2−2x−4=2(x2−x−2)
x2−x−2=x2−2x+x−2
x2−2x+x−2=x(x−2)+1(x−2)=(x+1)(x−2)
2(x2−x−2)=2(x+1)(x−2)
Vậy, đa thức \(x^{2} - 4 + x \left(\right. x - 2 \left.\right)\) được phân tích thành nhân tử là \(2 \left(\right. x + 1 \left.\right) \left(\right. x - 2 \left.\right)\).
\(x^2-4\) + \(x\left(x-2\right)\)
= (\(x^2-4\)) + \(x\left(x-2\right)\)
= \(\left(x+2\right)\left(x-2\right)\) + \(x\left(x-2\right)\)
= (\(x-2\))(\(x+2+x\))
= (\(x-2)\)((\(x+x)+2)\)
= (\(x-2)\)(2\(x+2\))
voãi
nah bruhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh