Phân tích đa thức thành nhân tử
\(x^3-2x^2+x-xy^2\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=x\left(x-4\right)+5\left(x-4\right)=\left(x+5\right)\left(x-4\right)\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
\(x^3-64=x^3-4^3\)
\(\Rightarrow\left(x-4\right)\left(x^2+4x+4^2\right)\)
x^2 + 7x -15
= x^2 + 7x +12,25 -27,25
= (x+3,5)^2 - 27, 25
= ( x+3,5 - \(\sqrt{27,25}\))(x+3,5+\(\sqrt{27,25}\))
Ta có : x5 + x + 1
= x5 + x4 - x4 - x3 + x3 + x2 - x2 - x + x + 1
= (x5 + x4) - (x4 + x3) + (x3 + x2) - (x2 + x) + (x + 1)
= x5(x + 1) - x4.(x + 1) + x3(x + 1) - x2(x + 1) + (x + 1)
= (x + 1)(x5 - x4 + x3 - x2 + 1)
\(x5+x-1 = x5-x4+x3+x4-x3+x2-x2+x-1 = x3(x2-x+1)+x2(x2-x+1)-(x2-x+1) = (x2-x+1)(x3+x2-1) \)
hc tốt nha !!!!!!!!!
\(x^5+x+1=x^5-x^2+x^2+x+1=x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1+y\right)\left(x-1-y\right)\)
= x (x2-2x+1-y2)
= x ( (x-1)2-y2)
=x (x-1-y) (x-1+y)